Plane and Solid Geometry Suggestive Method REVISED EDITION By George C. Shutts Instructor in Mathematics, State Normal School Whitewater, Wisconsin Atkinson, Mentzer & Company Boston New York Chicago Dallas Copyright, 1912 BY GEORGE C. SHUTTS TABLE OP CONTENTS Plane Geometry: Preface Introduction , 1 Axioms and Postulates 11 Symbols and Abbreviations 13 Rectilinear Figures Triangles 19 Quadrilaterals 52 Polygons 62 Locus 65 Circles 73 Construction Problems 87, 16? Measurement and Proportion 97 Trigonometric Relations 121 Measurement of Angles 126 Extreme and Mean Ratio 145 Numerical Relations 147 Area of Polygons 1 fi ") Regular Polygons 191 Incommensurable Magnitudes 21 3 Solid Geometry: Lines and Planes 223 Dihedral Angles 248 Polyhedral Angles 260 Polyhedrons 270 Prisms 271 Pyramids 290 Similar Polyhedrons 302 Regular Polyhedrons 306 Cylinders 309 Cones I 318 Spheres 330 Index . . .375 PREFACE. In this revision the suggestive features of former editions have been retained, and it is hoped improved. The principal value in the study of geometry lies in the power developed by the individual student in work- ing out as much as possible his own demonstrations. Pupils in their early study of the subject of geom- etry sometimes fail to see the need of some of the steps in a rigorous deductive demonstration and, to satisfy the demands of the class room, resort to formal memory of text as a substitute for logical thinking. This text has been prepared, not so much to illustrate a logical development of mathematical science as to arrange and adapt the mathematical data herein contained to the comprehension and growth of the pupil. In the intro"- duction of various subjects theorems are stated as postu- lates or preliminary propositions which in a strictly logical development would require proof. Many of these statements are of so fundamental a nature, have so long been accepted as obvious facts by the pupil, and yet are so difficult of demonstration, that the chief value in the use of them lies not so much in their demon- stration as in the comprehension and use of them in the demonstration of truths based upon them. Theo- rems have been chosen for the earlier demonstrations such that the course of reasoning will appeal to the logical ability of the pupils at that stage of their rea- soning power. ^59793 The protractor T square, parallel rulers/ etc., have been brought into use before the principles upon which their construction is based have been developed, that through a mechanical construction of the various forms, the pupil may get a better understanding of their mean- ing than he usually gets from abstract definitions. The rigor of the demonstrations has in no case been minimized and constructions, when the principles upon which they depend have been developed, as usual require the use of straight edge and dividers only. The subject of measurement and proportion has been simplified, and the incommensurable cases placed in the latter part of the text in plane geometry, with the intent that in a short course in plane geometry this phase may be omitted if the teacher deems it wise to do so. Many practical applications of the principles of geometry have been introduced in the exercises, but have been selected with reference to the interests of all the pupils, rather than the few who desire to enter technical occupations. Many of the old geometrical puz- zles have been omitted and numerical problems of in- terest introduced. It is not expected that any given class or pupil will work all of the exercises, but the number is large enough and the variety sufficiently great to furnish interesting recreation and drill for the var- iety of tastes in any class of students. Many exercises follow the propositions to which they are related for application and drill, but many others are scattered throughout the text more or less unrelated to serve the 1 purpose of review and test the pupil's power of indepen- dent thought. The one suggestion the author would make to the teacher is that success depends upon letlinir tin 1 nilc <>!' progress be determined by the thorough comprehension by the pupils of the subject matter. The amount of text covered is not so important as the real develop- ment of the pupils. The author wishes to acknowledge for their scholarly assistance, his obligation to H. D. Merrill of the chair of mathematics of the Evanston High School, who has made valuable suggestion throughout ; to A. W. Smith, Professor of Mathematics of Colgate University, Ham- ilton, N. Y., who assisted in the revision and read the proof, and to many students who through conscientious work in the class room have given valuable suggestions. To these old friends and to the prospective students, to whom it is hoped these pages will furnish inspiration, this work is dedicated. GEORGE C. SHUTTS. Whitewater, Wis., Aug. 1, 1912. PLANE GEOMETRY INTRODUCTION. 1. PHYSICAL SOLID. Any material object is a phys- ical solid, for example, a block of wood or a ball. 2. GEOMETRIC SOLID. A limited portion of space is a geometric solid. A block of wood is a physical solid but the portion of space occupied by it is a geometric* solid. Two reasons are suggested for this definition of a geometric solid: first, geometry is not concerned with the material of which an object is made but with its size and its shape ; second, such solids can be made to coincide in whole or in part. 3. SURFACE. That which has length and breadth without thickness is a surface. The boundaries cf a solid are surfaces but these surfaces are not a part of the solid which they bound any more than one's shadow on a wall is a part of the wall. A surface may pass through a solid or another surface and may be unlimited in extent. 4. LINE. That which has length without breadth or thickness is a line. A line may pass through a solid, a surface, or another line and may be unlimited in extent. 5. POINT. That which has position without dimen- sions is a point. 2 PLANE GEOMETRY 6. MAGNITUDES. Lines, surfaces, and solids are geometric magnitudes. 7. MAGNITUDES GENERATED. The path of a moving point is a line, or a point in motion generates a line. A line in motion (except upon itself) generates a surface. A surface in motion (except upon itself) generates a solid. 8. MAGNITUDES, LIMITED OR BOUNDED. Solids are limited or bounded by surfaces, surfaces by lines, lines by points, unless in accordance with the definitions given above they are unlimited. 9. GEOMETRIC FIGURES. Any combination of points, lines, surfaces, and solids is a geometric figure. Some Geometric figures are mental images only. They cannot be constructed and the drawings used are but represen- tations. For example, a point may be represented by a crayon or pencil dot, a line by a pencil mark or the edge of a ruler, a surface by a table top or the surface of a blackboard, etc. 10. STRAIGHT LINE. A line is straight if any part of it will lie wholly in any other part when its extrem- ities lie in that part, or if, when revolved with two points kept stationary, it occupies the same position as before. A straight line may be thought of as extending in- definitely in both directions. The first test may be ap- plied to a pencil mark representing a straight line by using tracing paper to apply "any part" to "any other part" of it or by putting together the edges of two rulers as representing two parts of a straight line. The second test may well be illustrated with a piece of straight wire. INTRODUCTION 3 11. LINE SEGMENT. A limited portion of a line is a line-segment. A line-segment is sometimes called a sect. 12. RAY. A portion of line limited at one end only is a ray. The limiting point of a ray is its origin. A segment differs from a ray in that it has two limiting points instead of one. 13. NOTATION. A point is read by naming a capital letter placed near it. A line is read by naming the let- ters which mark any two of its points. Lines, rays, and line-segments are at times indicated by single small letters instead of two capitals indicating points. For example in the following figures one reads : point A, line BC or I, segment DE or s, ray FG or r. 14. EQUAL SEGMENTS. If two line segments can be so placed that their extremities coincide the segments are said to be equal. 15. ADDITION AND SUBTRACTION OF SEGMENTS. (1) Two segments are added by placing them end to end. The resulting segment is called the sum. (2) Two segments are subtracted by placing the shorter upon the longer so that one pair of end points coincide. That portion of the longer not covered by the shorter is called the difference. For example, A B C D to find the sum A ^ D B- v of segment AB ~ ~~* and segment CD, draw an indefinite line and on it lay off A'B' equal to AB and B'D' equal to CD. Then A'D' is the sum of AB and CD or as it may be written in symbols, PLANE GEOMETRY AB + CD =-- A'D'. To find the difference between AB and CD lay CD upon AB with point C on A. The dif- ference is the segment DB. The relation is here written as AB CD = DB. The measurements here necessary may be made with a ruler or scale or better yet with a pair of dividers. 16. BROKEN LINE. A line made up of successive segments not form- ing a straight line is a broken line. 17. CURVED LINE. A line no part of which is straight is a curved line or simply a curve. 18. PLANE. A surface such that a straight line through any two of its points lies wholly in the surface is a plane. A plane is sometimes spoken of as a flat sur- face and portions of a plane may be enclosed by broken lines or curves. 19. PLANE FIGURES. A figure all parts of which lie in the same plane is a plane figure. (1) A figure all lines of which are straight is a rectilinear figure. (2) A portion of a plane entirely enclosed by a broken line is a plane polygon or simply a polygon. (3) A curve enclosing a portion of a plane in such a manner that every point of the curve is at the same dis- tance from a point within is a circle, the point within is the center, the distance from the center to the circle is the radius, and any portion of the circle is an arc. 20. PLANE GEOMETRY. That part of geometry which treats of plane figures is called plane geometry. 21. ANGLE. Two rays proceeding from the same point contain or make an angle. The rays are the sides INTRODUCTION 5 of the angle and the point of meeting is the vertex of the angle. An angle may be read by naming the letters which designate its sides, the vertex letter being read between the others, as the angle AOB. If no confusion results, an angle may be read by naming its vertex letter alone, Q^J^. A as the angle 0, or by naming a single small letter, as angle m. 22. TURNING A LINE THROUGH AN ANGLE. If a line coincident with one side of an angle be revolved about the vertex as a pivot until it coincides with the other side of the angle, the line turns through the angle. This turning is usually conceived to be counter-clockwise. 23. SIZE OP ANGLE. The size of an angle is the amount of revolution necessary to turn a line through it. It bears no relation to the length of the sides. 24. EQUAL ANGLES. If two angles can be placed so that their vertices coincide and the sides of the one lie upon the sides of the other the angles coincide and the turning through both angles is effected at the same time. Such angles are said to be equal. A pair of dividers may be used to represent an angle, the size of the angle depending upon the extent to which the dividers are opened. They may be used to compare two angles by placing them upon one of the angles with the legs of the dividers lying along the sides of the angle and then placing them upon the other angle, noticing whether or not it is necessary to open or close them in order that they fit the second angle. A similar compari- son may be made by drawing one angle on a piece of tracing paper and placing it upon the other angle so that the vertex and one side of the trace lie upon the vertex and one side of the second angle. The relative position of the remaining sides of the two angles show which is the larger. PLANE GEOMETRY 25. ADJACENT ANGLES. Two an- gles that have the same vertex and one common side between them are adjacent angles. The angles AOB and BOC are adjacent. 26. ADDITION AND SUBTRACTION OF ANGLES. (1) Two angles are added by so placing them as to form adjacent angles. The sum is the large angle thus formed. (2) Two angles are subtracted by placing the smaller within the larger so that they have one side and the vertex in common. The angle adja- cent to the smaller is the difference. For example, the sum of angles AOB and BOC is AOC and the difference between AOB and DOB is AOD. Are these results consistent with the statement in 22? By means of tracing paper obtain the sum of the angles AOB of 21 and BOC of 25. 27. PERIGON. The total angular magnitude about a point in a plane is a perigon. A line has turned through a perigon when it has made a complete ,""". _ * o ^ revolution about a point as BOB'. { 7fT~ 28. STRAIGHT ANGLE. An angle the sides of which lie in a straight line and on opposite sides of the vertex is a straight angle, as angle AOB. / ,*-- % A straight angle is one half a peri- B o "~A gon. The student must distinguish between the straight line AOB and the straight angle A OB. INTRODUCTION 7 Let the student turn a ray through the angle AOB. 29. RIGHT ANGLE. If a ray meets a line so that the two angles formed are equal, each angle is a right angle, as L AOB and LEOC. o (1) A right angle is one half of a straight angle. (2) A right angle is one fourth of a perigon. 30. DEGREE. One ninetieth part of a right angle is a degree. (1) One sixtieth part of a degree is a minute and one sixtieth part of a minute is a second. (2) Degrees, minutes, and seconds are denoted by the symbols , ', " respectively, as 14 27' 30". (3) The degree is a unit of measure for angles. A protractor is sometimes used in constructing and measuring angles. It consists of a semicircular scale with degrees from to 180 marked on it. To use the protractor in measuring an angle place the center at the vertex of the angle and the zero of the protractor upon one side of the angle. The number of degrees and minutes in the angle is the number indicated upon the protractor at the point where the second side of the angle projects beyond the protractor. The protractor can also be used in constructing an angle of a given number of degrees or one that is equal to a given angle. With the protractor construct a right angle, cut it out, and test the accuracy of the construction by superimposing it upon a known right angle. 8 PLANE GEOMETRY 31. ACUTE ANGLE. An angle less than a right angle is an acute angle. With a protractor construct acute angles of 20, 70, 65, 40, 28, 53', etc. 32. OBTUSE ANGLE. An angle greater than a right angle and less than a straight angle is an obtuse angle. With a protractor construct obtuse angles of 95, 115, 170. Add angles of 82 and 32; of 47 and 53. 33. REFLEX ANGLE. An angle greater than a straight angle and less i^ . than a perigon is a reflex angle. ^^ Reflex angles are seldom considered in elementary geometry. With a protractor or tracing paper add angles of 98 and 71 ; of 120 and 92; of 72 and 65. What kind of angles are the sums? 34. OBLIQUE ANGLES. Acute and obtuse angles are oblique angles. 35. COMPLEMENTS. Two angles the sum of which is a right angle are complements. With a protractor construct the complements of 60, 50, 75, 49. Measure each and test the results. 36. SUPPLEMENTS. Two angles the sum of which is a straight angle are supplements. The complement or supplement of a given angle need not be adjacent to it. Construct the supplement of 18, 24, 96, 125. Which are acute and which obtuse? Test the accuracy of the results with the protractor. 37. CONJUGATES. Two angles the sum of which is a perigon are conjugates. Two rays proceeding from a point really form two angles, as the angles ^ m and n. These angles are conju- /*"* N -^ gates. INTRODUCTION Measure the degrees in m and // and to^t the result by this definition. In referring to an angte the smaller of the two conju- gates is meant unless otherwise stated. The student should provide himself with a ruler, a protractor, a pair of dividers, parallel rulers, and if convenient a T square and drawing board. 1. In the adjacent figure AE is a straight line and CO is perpendicular to AE. Read a straight angle, two right angles, five acute angles, two ob- tuse angles. o~ 2. Eead ten pairs of adjacent angles in the above figure. 3. What is the sum of angles x and u? Of angles y and v? Of angles u and x? Of angles y and x? 4. Read two pairs of complementary angles in the above figure. 5. Read three pairs of supplementary angles in the above figure. 6. A perigou is equal to how many straight angles? A straight angle is equal to how many right angles! A perigon is equal to how many right angles? Compare a straight angle with a perigon ; a right angle with a perigon ; a right angle with a straight angle. 7. How many degrees are there in a straight angle? In a perigon? 8. How many degrees in the supplement of a right angle? In the supplement of two-thirds of a right angle? In the comple- ment of three-fourths of a right angle? 9. Find the supplement and the complement of 54 27' ; of 57 35' 42". 10. How many degrees in tha smaller angle formed by the hands of a clock at 1 o'clock? At 5 o'clock? How many degrees are generated by the minute hand of a clock from two o 'clock to twenty minutes past two? 11. A perigon is divided into six equal angles. How many degrees are there in each? Illustrate with a figure. 10 PLANE GEOMETRY 12. A perigon is divided into three angles, the second being twice the first arid the third three times the first. Find the num- ber of degrees in each angle. Illustrate with a figure. Suggestion: Let x equal the number of degrees in the first. 13. Write an equality that will say that angles x and y are complements; that they are supplements. 14. Angles x and y are supplements and their difference is 50. Find angle x and angle y. With a protractor draw x and y and check the results. 15. How can the supplement of an angle be found? Draw an angle, as x, with the protractor and find its supplement. 16. The supplement of an angle x is three times the comple- ment of x. Find the angle x. 17. In the figure of example 1 measure with a protractor the angles x, y, and z ; add them and test the result by measuring angle ADD. 18. Draw a straight angle. Divide it by a ray into two oblique angles. Measure each and add them. How many degrees should their sum be? 19. Divide a perigon into five angles and measure each. Test the accuracy of the work by adding the re- sults. 20. Measure angle A and angle CAD. Subtract the results and test the accuracy of the work by measuring angle DAB. 21. Draw a line-segment 1 inch long. Draw another 1$ inches long so as to include between them an angle of 45. Then measure the distance between the free ends of the segments. Make the construction a second time and with tracing paper test whether the two figures are congruent. 64. 22. Draw an angle of 25, 18, 72, 126, 175. 23. Draw a reflex angle of 197, 240, 315, 345. 38. VERTICAL ANGLES. When the sides of one angle are extensions of c the sides of another angle, the two angles are vertical angles. For example, m and n are vertical angles. INTRODUCTION 11 39. PERPENDICULAR LINES. If two lines form a right angle, each line is perpendicular to the other. The word perpendicular is used also as a noun. 40. OBLIQUE LINES. If two lines form an oblique angle, each line is oblique to the other. 41. BISECTOR. If a figure is divided into two equal parts it is bisected. In plane geometry the bisector of a segment may be a point, a segment, a ray, or a line. The bisector of an angle may be a segment, a ray, or a line. 42. PROOF. To prove a statement is to show by a logical course of reasoning that it must be true by means of other statements that have been accepted. In general the proof of a statement involves other statements which in turn require proof. Evidently as a foundation for reasonings some statements must be accepted without proof. 43. AXIOM. A statement admitted without proof to be true and not limited to geometric figures is an axiom. 44. POSTULATE. A statement admitted without proof and limited to geometric figures is a postulate. 45. THEOREM. A statement that is to be proved is a theorem. 46. PROBLEM. The statement of a geometric con- struction to be made is a problem. 47. PROPOSITION. Theorems and problems are prop- ositions. 48. COROLLARY. A proposition easily deduced from another proposition is a corollary. 49. AXIOMS. In the following axioms the equalities and inequalities concerned are unconditional. (1) Quantities that are equal to the same quantity, or to equal quantities, are equal to each other. 12 PLANE GEOMETRY (2) If equals are added to or subtracted from equals the results are equal. (3) If equals are multiplied or divided by equals (division by zero being excluded) the results are equal. (4) If equals are added to or subtracted from un- equals the results are unequal in the same order. (5) If unequals are multiplied or divided by posi- tive equals the results are unequal in the same order. (6) If unequals are added to unequals in the same order the results are unequal in that order. (7) If unequals are subtracted from equals the re- sults are unequal in the reverse order. (8) If the first of three quantities is greater than the second and the second is greater than the third, then the first is greater than the third. (9) In considerations involving size only, the whole is greater than any of its parts and is always equal to the sum of its parts. (10) Either of two equals may be substituted for the other in any process. (11) Every magnitude, however small, can be divided into two or more parts. 50. POSTULATES. (1) If two straight lines have two points in com- mon they become one and the same straight line. Between two points there is one and only one straight line. (2) The shortest path between two points is the straight line segment joining them. (3) A line-segment has one and but one mid-point. An angle has one and only one bisector. (4) A figure can be moved without altering its size or its shape. (5) A line in a plane divides the points of that plane INTRODUCTION 13 into two classes such that a line-segment connecting two points of the same class is not intersected by the line and a line-segment connecting two points not of the same group is intersected by the line. (6) Magnitudes which coincide are equal. (7) All perigons are equal. (8) A circle or an arc can be constructed with any radius and any given point as center. (9) A straight line can be drawn between two points ; a straight line can be terminated at any point ; a sect can be extended any distance. SYMBOLS AND ABBREVIATIONS. The symbols +, , X > is greater than. < is less than. 1 perpendicular. I! parallel. = is equal to. '-' is similar to. ~ is congruent to. m is measured by. = approaches as a limit. L angle. A triangle. O parallelogram. Adj. adjacent. Alt. alternate. Auth. authority. Ax. axiom. Const, construction. Cor. corollary. ^~ are used as in algebra. Def. definition. Ex. exercise. Ext. exterior. Fig. figure. Hyp. hypothesis. O rectangle. O circle. article, rt. right, st. straight. " therefore. ** hence. Iden. identity: Int. interior. Post, postulate. Prop, proposition. Sug. suggestion. Sup. supplementary. 14 PLANE GEOMETRY The symbols Z, A, O, CD, O have the plural forms A, A, O7, m, . PRELIMINARY THEOREMS. The following are some im- portant preliminary theorems : 51. THEOREM. 1. Straight angles are equal. SUG. 28, Post. 7, and Ax. 3. 52. THEOREM 2. Right angles are equal. SUG. 29 and Ax. 3. 53. THEOREM 3. (a) Complements of equal angles are equal, (b) Supplements of equal angles are equal. a. Given angles A and B, each being a complement of Z C. To prove Z A and Z 5 equal. SUG. 1. How does the sum A of A A and C compare with the sum of A. B and (7? Th. 2 and Ax. 1. 2. Compare Z A and Z B. Ax. 2. Therefore b. Follow the method of demonstration used for theo- rem 3 (a). N te. Before looking up references in the suggestions the pupil should endeavor to determine the authorities for himself. 54. THEOREM 4. Two right angles are supplements of each other. 55. THEOREM 5. Only one line can be drawn perpendicular to a line at a given point on the line. SUG. Can both OD and OC be 1 to AB at point 0? 29, 39 * and 50 (3). 56. THEOREM 6. Two lines that intersect can have but one point in common. INTRODUCTION 15 SUG. If they had two points in common, what would be true ? Post. 1. 57. THEOREM 7. // two straight lines intersect the vertical angles are equal. Given AB and CD intersect- ing at with vertical angles in and p. To Prove Z m = Z p. SUG. 1. What relation does Z m bear to Z n? 36. 2. What relation does L p bear to Z ?i ? . 36. 3. Compare Z m and L p. 53 (7J. Therefore Compare angles q and w. 1. Draw two supplementary adjacent angles and bisect each. How many degrees in the angle of the bisectors? Use the pro- tractor and straight edge. 2. How many degrees in the angle formed by the bisectors of two complementary adjacent angles? By the bisectors of two adjacent angles of 63 ari 70 respectively? Of 26 and 84 respectively? 3. If two lines intersect and one of the angles is a right angle, prove the others are right angles. 4. If one angle formed by two intersecting lines is 70 how many degrees are there in each of the others? If one angle is 50 how many in each of the others? 5. The difference between two complementary angles is 12. What are the angles? If the angles are supplementary, what are they? 6. Solve problem 5 if the difference is 40 ; 75 ; 82 16'. 7. The ratio of two complementary angles is 4:5. What are the angles? Solve for ratios of 3:6, 3^:5^, 5:9. 8. Solve problem 7 with the supposition that the angles are supplementary. 16 PLANE GEOMETRY 9. What is the complement of 100? The supplement of 212? Of 198 27'? If the student by use of algebra derives these results from the definitions of complements and supple- ments, he will have angles with the negative sign. The interpre- tation is this: Angles which are generated by counterclockwise notation are called positive and those generated by a clockwise notation are called negative. This plan is of great use whenever it is necessary to distinguish between these two ways of generat- ing an angle. 58. SUMMARY. 1. Two angles are equal (1) if they can be made to coincide; (2) if they are equal to the same angle or to equal angles; , (3) if they can be obtained by adding equal an- gles to equal angles or by subtracting equal angles from equal angles ; (4) if they are doubles or halves of the same angle or of equal angles; (5) if they are straight angles, right angles, or vertical angles; (6) if they are complements or supplements of the same or of equal angles. 2. Two angles are complements if their sum is a right angle or 90. 3. Two angles are supplements if their sum is n straight angle, two right angles, or 180. 4. An angle is a right angle (1) if it is one of two equal adjacent angles formed by two straight lines; (2) if it is one half of a straight angle or con- tains 90 ; (3) if its sides are perpendicular to each other; INTRODUCTION 17 (4) if it is one of two equal supplementary an- gles. 5. An angle is a straight angle ( 1 ) if its sides form a straight line ; (2) if it is the sum of two right angles or 180 ; (3) if it is one-half of a perigon. 6. Two lines are perpendicular to each other if they form a right angle. 7. Two lines form one straight line (1) if they have two points in common; (2) if they are the sides of a straight angle; (3) if they are perpendicular to the same line at the same point. 8. Line segments or sects can be proved equal by means of axioms 1, 2, 3 or by showing that they can be so placed that their end points coincide. 9. Angles or line segments can be proved unequal by means of axioms 4-9 inclusive or postulate 3. . ' - .'<. >Y> i i 2> -i CHAPTER I. RECTILINEAR FIGURES. 59. TRIANGLE. A portion of a plane bounded by three sects is a triangle. The line-segments are the sides of the triangle, their intersections are the vertices, their sum is the perimeter, the angles formed by the sides are the angles of the triangle, and the three sides and three angles are the parts of the triangle. 60. TRIANGLES CLASSIFIED AS TO SIDES. A triangle no two of the sides of which are equal is a scalene triangle. One with two equal sides is an isosceles triangle. One with three equal sides is an equilateral triangle. In the accompanying figures A ABC is scalene, A DEF is isosceles, and A GHI is equilateral. Measure to verify. C F I 61. TRIANGLES CLASSIFIED AS TO ANGLES. A triangle one of the angles of which is a right angle is a right triangle. One with an obtuse angle is an obtuse trian- gle. If all the angles are acute the triangle is an acute triangle. If all the angles are equal the triangle *is equiangular. A ABC is right, A DEF is obtuse, A GHI is acute. Verify with a protractor. 20 PLANE GEOMETRY 62. BASE OF A TRIANGLE. Unless otherwise stated, the base of an isosceles triangle is the side which is not one .of the two equal sides. In other cases the base of a triangle is that side upon which it is conceived to stand. 63. VERTEX ANGLE OF A TRIANGLE. The angle op- posite the base of a triangle is the vertex angle and the vertex of this angle is the vertex of the triangle. 64. CONGRUENT FIGURES. Two figures that can be made to coincide throughout are congruent. Congru- ent means of the same shape and size, while equal means of the same size only. 1. Draw a triangle with two sides 2 and 2| inches respect- ively and included angle 75. [30, (3)]. Draw another with the same conditions. Use tracing paper to compare them or cut one out and compare them by superposition. Are they congruent? 2. Draw a triangle at random. Measure two sides and the included angle. Construct another using the measurements ob- tained from the first and compare them by superposition. 3. Draw a triangle with two angles of 85 and 65 respec- tively with the included side 2 inches. Draw another with the same conditions. Compare them by superposition. What is the conclusion? 4. Draw a triangle at random. Measure two angles and the included side. Construct another triangle from these measure- ments and compare them by superposition. 5. Draw a straight line "1|- inches long. With the dividers take lines respectively f and 1 inches and from the ends of the first line describe arcs that intersect. Connect the point of intersection with the ends of the first line. Take the same three lines in any order and construct a triangle. Compare these tri- angles as regards congruence. 6. Construct a triangle in which the sides are 2, 3, and 4 inches respectively. Construct a triangle with lines of l, If and 4 inches respectively. Try again with lines of 2, 2, and 4 inches respectively ; 2, 3, and 7 ; 4, 5, and 6. 7. State the conclusions of Ex. 6 as to the possibility of con- structing a triangle of given sides. RECTILINEAR FIGURES 21 PROPOSITION I. 65. THEOREM. Two triangles are congruent if two sides and the included angle of the one are equal respectively to two sides and the included angle of the other. A A Given A ABC and & A'B'C' with BA B'A f , BC = B'C', and Z B= L B' . To Prove A ABC = A A'B'C'. Proof . Place A ABC upon A A'B'C 1 so that point B will fall on B' and BC will lie along B'C'. Then C will fall on C' for BC = B'C'. BA will fall along B' A' for L B = L B'. A will fall on A" for BA = B'A', Since C falls on C' and J. falls on A', line AC will coincide with A'C' for only one straight line can be drawn through two given points. ' A ABC = A A'B'C' for they can be made to coincide. Therefore, Two triangles are congruent if two sides and the included angle of the one are equal respectively to two sides and the included angle of the other. 1. To find the distance across an impassable marsh: Stake off two lines AC and BD that cross at an accessible point 0. Measure distances OD equal to OA and OB equal to OB. Then the required distance AB is equal to CD. Why? _ 2. In the school grounds or elsewhere drive three stakes for A, B, and of ex- 22 PLANE GEOMETRY ample 1. Make the remaining measurements and test the ac. curacy of the work by comparing AB and CD. PROPOSITION II. 66. THEOREM. Two triangles are congruent if two angles and the included side of the one are equal to two angles and the included side respec- tively of the other. Given A ABC and A A' B'C' with /.B=/.B', BC = B'C', LC= L C'. To Prove A ABC = A A'B'C'. Proof. Place A ABC upon A A'B'C' so that B shall fall on B r and BC shall fall along B'C'. Then C will fall on C' for BC is given equal to B'C'. BA will fall along B'A' for L B= L B' . '-A will fall on B'A' or B'A' produced. Also CA will fall along C'A' for L C= L C'. '-A will fall on C'A' or C'A' produced. ' A must fall on A', the intersection of B'A' and C'A'. /WhyV) ' A^tgtT^ A A'B'C' for they can be made to coin- cide. Therefore, Two triangles are congruent if two angles and the included side of the one are equal to two angles and the included side respectively of the other. 67. APPLICATION OP PROPOSITIONS I AND II. If two triangles are congruent the angles are equal respectively RECTILINEAR FIGURES 23 and the sides are equal respectively. It is important to keep this in mind for usually the purpose in proving triangles congruent is to prove that a pair of angles or line-segments are equal. In congruent triangles pairs of equal sides lie opposite pairs of equal angles and pairs of equal angles lie opposite pairs of equal sides. The pairs of equal parts are called corresponding or homolo- gous parts. 1. Draw a triangle at random and with a straight edge and protractor; construct a second triangle having two sides and their included angle of the same dimensions as the corresponding parts of the first. Why must the triangles be congruent? Compare the two, using tracing paper, as a test for accuracy of construction. PROPOSITION III. 68. THEOREM. // two sides of a triangle are equal, the angles opposite those sides are equal. Given A ABC with AB = AC. To Prove Z B = Z C. Proof. Draw AM to represent the bisector of Z A and extend it to meet B( 1 as at M, A. B and C will be. proved equal if it can be shown that A AMB and AMC are congruent. A comparison of the two triangles shows AB = AC by hyp. Z MAB = L MAC since AM bisects Z A. AM is common to the two triangles. 'A AMB = A AMC, two sides and the in- cluded angle of the one being equal to two sides and the included angle of the other. Z B is in A AMB and is opposite to side AM. Z C is in A AMC and is opposite side AM. 24 PLANE GEOMETRY ' Z B = L C, being homologous angles in con- gruent triangles. Therefore, If two sides of a triangle are equal, the angles opposite those sides are equal. REMARK. Observe that LB is an angle in each of two tri- angles. In A AMB it lies opposite the side AM, in A ABC it lies opposite side AC. Likewise LC is an angle in each of two triangles. 69. COROLLARY. An equilateral triangle is equiangu- lar. Proof. Use proposition III. PROPOSITION IV. 70. THEOREM. The bisector of the vertex an- gle of an isosceles triangle bisects the base and is perpendicular to the base. ** C Given A ABC, with AB = AC and Z MAB = Z MAC. To Prove BM = MC and AM LBC. Proof. A comparison of the A AMB and A AMC shows AB = AC by hyp. Z MAB = MAC by hyp. Z B Z (7, being opposite equal sides of the isosceles A ABC. 'A AMB E A AMC, having two angles and the included side of the one equal to the cor- responding parts of the other. RECTILINEAR FIGURES 25 In A AMB side BM is opposite Z MAB and in A MAC side M.C is opposite L MAC. -'BM = MC, being homologous sides in con- gruent triangles. Also Z AMB = Z AMC, being homologous an- gles in congruent triangles. 'A. AMB and AMC are right angles. Why? 'AMBC. Why? Therefore, The bisector of the vertex angle of an isosceles triangle bisects the base and is perpendicular to the base. 71. DISTANCE. The length of the sect joining two points is the distance between these points. It is to be noted that distance is always measured on the shortest line. 72. PERPENDICULAR BISECTOR. A perpendicular erected at the mid-point of a line-segment is the per- pendicular bisector of that segment, PROPOSITION V. Z3. THEOREM. Any point in the perpendicular bisector of a line-segment is equidistant from the extremities of the segment. B ^ M O Given segment BC with BM MC, MN 1 BC and A any point in MN. To Prove AB = AC. 26 PLANE GEOMETRY Proof. SUG. If A BMA and A CM A can be proved congruent, segments AB and AC will be seen to be equal. Search the conditions given to see if there are enough elements equal to make the tri- angles congruent by Prop. I or II. See method of 70. Compare AB and AC, giving authorities. 74. PARTS OF A THEOREM. From the propositions studied thus far it is seen that in a theorem certain con- ditions are given and a certain thing is to be proved. The conditions given constitute the hypothesis and the truth to be established is the conclusion. The course of reasoning by means of which the conclusion is estab- lished is the proof or demonstration. Each statement in a proof must ,be justified by reference to the hypoth- esis or to definition, axiom, postulate, or former prop- osition. 1. Two right triangles are congruent if the sides of the right angle of the one are equal respectively to the sides of the right angle of the other. 2. Lines AB and CD bisect each other at 0. Prove A AOC = A BOD and AC = BD. o 3. OC bisects L AOB and DELOC at F. Prove A OFD - A OFE, OD = OE, and L ODF - Z OEF. 4. In A ABC, AB = AC and Z BAM = L CAN. Prove A BAM = A CAN and &AMN is isosceles. 5. In A ABC, AB = AC and D, E, F are the mid-points of AB, BC, and CA re- spectively. Prove A DEF isosceles. RECTILINEAR FIGURES 27 PROPOSITION VI. 75. THEOREM. Two triangles are congruent if the sides of the one are equal respectively to the sides of the other. Given A ABC and A A'B'C', AB = A'B', BC = B'C and CA = C' A' . To Prove A ABC E A A'B'C'. SUG. 1. The conclusion follows at once if it can be shown that Z A = L A', Z5=Z5', or Z C = L C'. 2. Place A A'B'C' so that a side, as A'C' coincides with its equal AC, the two points B and B' being on opposite sides of AC. 3. Draw BB'. 4. Z ABB' = L AB'B. Why ? 5. Z 055 ' = Z 05 ' 5. Why ? 6. ' Z ABC = Z A5'C. Why ? 7. ' A A50 = A A'B'C'. Why? Demonstrate the theorem again, placing a second pair of equal sides in coincidence. Therefore 1. The straight lines which bisect the equal angles of an isosceles triangle and terminate in the opposite sides are equal. 2. A ABC is equilateral and AD = BE = CF. Prove A DEF equilateral. 28 PLANE GEOMETRY PROPOSITION VII. 76. THEOKEM. Any point ivhich is equidistant from the extremities of a line-segment is in the perpendicular bisector of the segment. The figure is to be constructed by the pupil. Given the segment AB with mid-point C, and D any point such that DA DB and ME the line through C and D. To Prove ME 1 AB at C. SUG. Compare & ACD and BCD and also the two adjacent angles at C. 77. COROLLARY. Any two points each equidistant from the extremities of a line-segment determine the per- pendicular bisector of the segment. SUG. How many points determine a line? 50 (1). 1. Nail together with one nail at each vertex three narrow strips of wood so as to form a triangle. Can it be racked out of shape without loosening the joints? Why? . 2. Remove the bottom from a chalk box. Can the box now be racked out of shape? Nail a strip across two adjacent sides. Can it still be racked out of shape? Give authorities for your answers. 3. Why does the stay-lath AB which the carpenter nails across a window or door frame hold it in a fixed shape? 4. Observe at the next opportunity the board nailed obliquely across the studding as they are erected in putting up the frame of a building or the diagonal strip on a gate, as AB in the figure. 5. Name at least five other applications of prop. VI for insuring stability in mechanical constructions. 6. If the perpendicular bisector of the base of a triangle passes through the vertex, the triangle is isosceles. Use fig. of 8 70. RECTILINEAR FIGURES 29 PROPOSITION VIII. 78. PROBLEM. Construct a triangle when the three sides are given. ^ Given sides of a triangle as a, 6, c. To Construct the triangle. SUG. 1. Draw a line and with dividers lay off upon it one of the sides as AB = c. 2. With the dividers find a point C that is the distance a from B and the distance 6 from A. 50 (8). A ABC is the required triangle. PROPOSITION IX. 79. PROBLEM. Construct upon a given ray an angle equal to a given angle. E Given ray AB and ZC. To Construct an Z upon AB equal to Z C with its vertex at A. SUG. 1. With dividers measure equal dis- tances as CD and CE on the sides of Z C. 2. Find F on AB so that AF = CE. 3. With the dividers find point X so that AX = CD and FX = ED. 4. Then Z XAF= Z DCE or Z C. For the 4 AXF and CDE have the three sides re- spectively equal. 30 PLANE GEOMETRY PROPOSITION X. 80. PROBLEM. Construct a triangle congruent to a given triangle. Given A ABC. A To Construct a A congruent to A ABC. Make three constructions. SUG. 1. Use the problem 78. 2. Use 65. 3. Use 66. In these problems use a straight edge and dividers only. 1. Prove the theorem of 75, using two scalene triangles with a pair of the shorter sides coincident. 2. Upon a given segment as a base construct an equilateral triangle. 3. Construct the perpendicular bisector of a given line-seg- ment. Sug. 77. P 4. Let be a given point in AB. Construct the perpendicular to AB through 0. Sue. If P be a point in this perpendicular, how may P be obtained? Why must line OP when thus determined be the required perpendicular. o< Solve example 4 assuming that O is outside AB. 6. Required to bisect a given an- gle, Z ABC. Sue. Obtain a point P such that when BP is drawn, Z DBF = Z EBP. 75. A 7. To measure a distance AB across a river. From A take any convenient line, AC, of known length, measure angles A and C. In some convenient place construct a C 1 A congruent to A ABC and measure the side corresponding to AB. State the authority for the conclusion. Sug. $ 60. RECTILINEAR FIGURES 31 PROPOSITION XL 81. THEOREM. Only one perpendicular can be drawn from a given point to a given line. M\ \ \ \ \ Given line CD with A a point not in CD, AO 1 CD, and AM any other line from A to CD. To Prove AM not perpendicular to CD. SUG. 1. Produce AO to B, making OB = OA. Connect M and B. 2. Since AOB is a straight line by con- struction, AMB is not a straight line. Why? 3. * Z AMB is not a straight angle. 4. Compare A AOM and A BOM. Auth. 5. Compare Z ^.ifO with Z 53fO. 6. ' Z ^MO is one half of Z AM. 7. ' Z J.MO cannot be a right angle. 8. ' AM is not perpendicular to CD. 9. ' AO is the only perpendicular from A to CD, since J.M represented all other lines from A to CD. 82. HYPOTENUSE. The side of a right triangle opposite the right angle is the hypotenuse. The other sides are usually called the sides or legs. 1 . Upon a given segment as a base construct an isosceles tri- angle. 32 PLANE GEOMETRY PROPOSITION XII. 83. THEOREM. Two right triangles are con- gruent if the hypothenuse and an adjacent angle of the one are equal respectively to the hypothe- nuse and an adjacent angle of the other. Given A ABC and A A'B'C' with L B and LE' right angles, AC = A'C', and Z C=Z C". To Prove A ABC = A A'B'C'. Proof. Place A ABC upon A A'B'C' so that A falls on A and AC falls along A'C". Where will C fall and why ? Also C will fall along C'B' for Z C= L C'. B will fall on C'B f or C"' produced. Why? Since A falls on A' and <7 falls along B'C', AB is the perpendicular from A' to B'C". But A'B' is by hyp. perpendicular to B'C'. Why? AB must fall along A'B'. Why ? '5 falls on B'. Why? A ABC = A A'B'C'. Why? Therefore 1. If two isosceles triangles have their bases coincident and their vertices on opposite sides of the common base, prove: (1) The entire figure is divided into two congruent tri- angles by a line connecting the opposite vertices. (2) This line is the perpendicular bisector of the common base. 2. If in example 1 the vertices are assumed to be on the same side of the common base, are the above conclusions true? RECTILINEAR FIGURES 33 PROPOSITION XIII. 84. THEOREM. Two right triangles are con- gruent if the hypothenuse and a side of the one are equal respectively to the hypothenuse and a side of the other. ^C B Z B' pro- Given A ABC and AA'B'C', with Z5 and right angles, AC = A'C', and AB = A'B'. To Prove A ABC = A A'B'C'. Proof. SUG. 1. The triangles are congruent vided Z C= /. C' or /. A= A' . 83. 2. Place the triangles so that A' falls on J., A'R' along J.#, and C and C' on opposite sides of AB. 3. Where will B' fall and why? OBC' is a straight line. Why? Figure ACC' is a triangle. Why? What kind of a triangle is ACC' ? Why? 7. Compare Z C and Z C". Auth. 8. Compare A ABC and J/5'C". Auth. Therefore Yv r hy is it necessary to prove CBC' a straight line? 85. COR. The perpendicular from the vertex to the base of an isosceles triangle bisects the ~base and the ver- tex angle. 4. 5. 6. 34 PLANE GEOMETRY 86. PARALLEL LINES. Straight lines that lie in the same plane and cannot meet however far produced are parallel lines. 87. POSTULATE. Through a given point only one line can be drawn parallel to a given straight line. PROPOSITION XIV. 88. THEOREM. Two lines in the same plane and perpendicular to the same line are parallel. M Given AB and CD in the same plane and each per- pendicular to MN. To Prove AB II CD. Proof. Suppose AB and CD ngt parallel, i.e. sup- pose they meet if sufficiently produced. Then from this point of meeting there will be two perpendiculars to the line MN, viz. AB and CD. But this is impossible for only one perpendicular can be drawn from a given point to a given line. 81. Hence the supposition that AB and CD meet is false. ' AB || CD, for they are in the same plane and do not meet. Therefore 89. INDIRECT PROOF. The proof in 88 is an example of indirect proof. In this method the theorem is assumed to be not true and the assumption is then shown to be false in that it leads to the contradiction of known facts. 90 is another example of indirect proof. RECTILINEAR FIGURES 35 1. The perpendiculars from the extremities of the base of an isosceles triangle to the opposite sides are equal. 2. Discuss exercise 1, when the vertex angle is obtuse; when it is right. 3. The perpendiculars drawn from the mid-point of the base of an isosceles triangle to the other sides are equal. 4. In &ABC, D is the mid-point of BC. BE and CF are perpendiculars drawn from B and C respectively to AD (AD be- ing produced through D). Prove BE = CF. 5. Discuss exercise 4, if A ABC is isosceles with AB = AC. PROPOSITION XV. 90. THEOREM. // one of two parallel lines is perpendicular to a given line f the other is per- pendicular to the same line. ~~G N Given AB \\ CD, AB 1 MN at E and CD intersecting MN at F. To Prove CD 1 MN. Proof. Suppose that CD is not perpendicular to MN. Let HGr represent the line which is perpendicular to MN, at F. Then HGllAB. Why? But CD II AB. Hyp. Then through F there are two lines parallel to AB. This contradicts the postulate, 87. ' the supposition that CD is not perpendicular to MN is false. -'-CD1.MN. Therefore 36 PLANE GEOMETRY 91. DISCUSSION. Observe that HG is given the prop- erty which CD is assumed not to possess. This is legiti- mate since it is known that through point F there is a line perpendicular to MN and if CD is not this perpen- dicular we may represent it by HG. DIRECT PROOF. Without making any supposition as to CD, draw a perpendicular to MN through F. Then as before both HG and CD are parallel to AB. Whence HG, if correctly drawn, must coincide with CD for other- wise there would be two lines through a given point parallel to the same line. Since HG 1 MN by construction and CD coincides with H G, then CD 1 MN. 92. TRANSVERSAL. A line that cuts two or more lines is a transversal or secant line. 93. CLASSIFICATION OF ANGLES MADE BY A TRANS- VERSAL. The angles made by two lines and a trans- versal are classified as follows: 1. The angles 1, 2, 3 and 4 are interior angles. 2. The angles 5, 6, 7 and 8 are exterior angles. c- 3. The pairs 1, 4 and 2, 3 are al- ternate-interior angles. F 4. The pairs 5, 8 and 6, 7 are alternate-exterior angles. 5. The pairs (1, 7), (2, 8), (6, 4), and (5, 3) are corresponding anglesW 1. Any point in the bisector of an angle is equidistant from the sides of the angle. ^^ Suo. Draw perpendiculars from a point l^me bisector to the si^s apd prove them equal. RECTILINEAR FIGURES 37 PROPOSITION XVI. 94. THEOREM. // two parallel lines are cut by a transversal the alternate-interior angles are equal. M G ^* p F' Given AB I! CD and cut by the transversal EF at points G and H respectively. To Prove Z AGF= Z DHE and Z BGF= CHE. SUG. 1. Through 0, the mid-point of GH, draw ON 1 CD and extend it to AB at M. 2. What relation does J/N sustain to AB and why? 3. Compare A OMG and A ONH. Auth. 4. Compare /. AGF and Z Z>##. 5. Compare Z <7,F and Z CHE. Therefore REMARK. In answering Sug. 4 observe that it is not known that OM and ON are equal. How then can it be shown that Z AGF and Z. DHE are homologous angles in congruent triangles? 1. Any point equidistant from the sides of an angle is in the bisector of the angle. 2. Any point not in the bisector of an angle is not equidistant from the sides. Sue. Draw PM 1 BC and PN 1 BA. Let PN meet the bisector in 0. Drato OG1.BC. Then OG = ON. Why? PG < PO + OG. ' PG < PN. PM < PG. ' PM < PN. Give the i^^n for each of these steps. 3. Demonstrate Ex. 2 by the indirect method. ^ 38 PLANE GEOMETRY The pupil is expected to make detailed proofs of the following three corollaries. 95. COR. // two parallel lines are cut by a trans- versal, the corresponding angles are equal. 96. COR. // two parallel lines are cut by a trans- versal, the alternate-exterior angles are equal. 97. COR. // two parallel lines are cut by a trans- versal, the interior angles on the same side of the trans- versal are supplementary. PROPOSITION XVII. 98. THEOKEM. // two lines in the same plane are cut by a transversal and the alternate-interior angles are equal, the lines are parallel. Given AB and CD cut by the transversal EF at G and H respectively with Z AGF = L DEE. To Prove AB II CD. SUG. 1. Suppose AB not parallel to CD and let MN represent that line through G which is parallel to CD. 2. Compare Z MGF with Z DHE. Auth. 3. Compare Z AGF with Z DUE. Auth. 4. Compare Z MGF with Z AGF. Auth. 5. Compare this conclusion with the facts 6. What of the supposition that AB is not parallel to CD? 1. What is the true relation between AB and CD? EECTILINEAR FIGURES 39 Therefore Prove this proposition by the direct method. See 91. REMARK. The second of the above figures will be useful if chalks of different colors are used to represent the lines AB and MN. 99. COR. I. // two lines in the same plane are cut by a transversal and the corresponding angles are equal, the lines are parallel. 100. COR. II. // two line in the same plane are cut l)y a transversal and the interior angles on the same side of the transversal are supplementary, the lines are parallel. 101. CONVERSE. If two propositions are so related that a condition of the first is the conclusion of the sec- ond and the conclusion of the first is a condition of the second, each proposition is called the converse of the other. For example, Prop. XVII is the converse of Prop. XVI for in the first of these the hypothesis is ^1 B II CD and the conclusion is Z AGF= Z DEE; in the second the hypothesis is Z AGF L DEE and the conclusion is AB II CD. It is to be noted that there is one condition common to the two propositions, viz: two lines and a transversal are given : The converse of a true statement is not always true. For example, all right angles are equal but not all equal angles are right angles. 1. If AB\\CD and x = 70, how many A / g degrees in each one of the various angles in y/ the figure? ./ T> 2. If AB || CD and y x = 80, how many degrees in each of the various angles? How many if s rl(f JL.f .'^f : Slff 7835 40 PLANE GEOMETRY 3. Two angles are complementary and their difference is 38. How many degrees in each? 4. Given a line AB and a point C not on AB. Construct a line through C and parallel to AB. Sug. Try to reproduce the conditions of Prop. XVII or of one of its corollaries. PROPOSITION XVIII. 102. THEOREM. Two angles the sides of which are respectively parallel and extend in the same or in opposite directions from the vertices are equal. Given A A, B, B' with AC I! BE, AD II BF and ex- tending in the same directions from the vertices A and B; and with AC il B'E', AD II B'F' and extending in opposite directions from the vertices A and B'. To Prove L A = L B = L B'. SUG. 1. Extend AD and B'E' until they meet at 0. (Why must they meet?) 2. Compare A A and B' '. 3. Extend AD and BE until they meet at M. 4. Compare /i A and B. Therefore- in what kind of triangles does the bisector of the vertex angle coincide with the perpendicular from the vertex to the base and with the median to the base I The line from the vertex to the mid-point of the base is called the median to the base. RECTILINEAR FIGURES 41 103. EXTERIOR ANGLE. An angle formed A B by one side of a rectilinear figure and an ^ Ss v adjacent side extended is an exterior angle. In the figure n is an exterior angle of the triangle ABC. 104. OPPOSITE INTERIOR ANGLE. In the above figure with respect to the exterior angle n, A and C are the opposite interior angles. PROPOSITION XIX. 105. THEOREM. An exterior angle of a trian- gle is equal to the sum of the opposite interior angles. *- c Given. A ABC with exterior angle DAC. To Prove Z DAC= Z B + L C. SUG. 1. Through the vertex A draw line MN \\ BC. 2. Compare Z p with Z B. Auth. 3. Compare Z q with Z C. Auth. 4. Complete the proof. Therefore 1. Two lines in the same plane, each parallel to a third line in that plane, are parallel. Sug. Draw a transversal, a line per- pendicular to the third line, or use the indirect method. 2. Two lines perpendicular to two intersecting lines respec- tively cannot be parallel. Sug. Use indirect proof. 3. Two angles are supplementary and their difference is 65. Find the angles. 4. State all methods thus far given for proving two lines parallel. 42 PLANE GEOMETRY PROPOSITION XX. 106. THEOKEM. The sum of the interior an- gles of a triangle is equal to two right angles. D * Given A ABC. To Prove ZA + Z+ZC = 2rt. A. Sue. 1. Extend one side as CB. 2. Compare Z m + Z n with Z , + Z 4 + Z C. Therefore 107. COR. A triangle can have at most one obtuse or one right angle. 108. COR. The acute angles of a right triangle are complementary. 1. Draw through B of the figure of 105 a line parallel to AC and prove the theorem. 2. Extend side CA through A to a point E and prove the theorem using the angles p, EAD, RAM. 3. Make the constructions of exercises 1 and 2 for exterior angles at B and C and prove the theorem. 4. The bisector of one of two vertical angles is the bisector of the other. 5. The bisectors of two vertical angles are in the same straight line. Sug. Prove that the bisectors form a straight angle. 0. The bisectors of two supplementary adjacent angles an- perpendicular to eacn other. RECTILINEAR FIGURES 43 PKOPOSITION XXI. 109. THEOREM. // two triangles have two an- gles of the one equal respectively to two angles of the other, the third angles are equal. Given A ABC and A A'E'C' with L A = L A' and Z=Z', To Prove ZC=ZC'. SUG. Use 106. 110. COR. // two right triangles have an acute angle of the one equal to an acute angle of the other, the re- maining angles are equal. 1. How many degrees in each angle of an equilateral triangle? 2. Construct an angle of 60; an angle of 30; an angle of 45. Use dividers and straight edge. 3. Construct a right triangle with acute angles of 60 and 30. 4. Trisect a right angle. As a general problem the trisection of an angle is impossible by the use of dividers and straight edge only. The pupil should note that this example is a particular case of the unsolved general problem. 5. How many degrees in each angle at the base of an isosceles right triangle? 6. The vertex angle of an isosceles triangle is 34 40'. How many degrees in each angle at the base of the triangle! 7. An exterior angle at the base of an isosceles triangle is 100. Find each angle of the triangle. 8. The angles of a triangle are 5z, 25x, and SO . Find the unknown angles. 9. The acute angles of a right triangle are x and 8#. Find them. 10. The difference between the acute angles of a right tri- angle is 26. Find them. 11. An exterior angle to one of the acute angles of a right triangle is 140. Find the angles of the triangle. 44 PLANE GEOMETRY PROPOSITION XXII. 111. THEOREM. // two angles of a triangle are equal, the sides opposite are equal and the triangle is isosceles. Given A ABC with L B = L C. To Prove AB = AC. SUG. as AM. Auth. Auth. 1. Drop a perpendicular from A to BC, 2. Compare Z BAM and L CAM. 3. Compare A BAM and A CAM. 4. Compare AB and AC. Therefore 1. The vertex angle of an isosceles triangle is i o: ? an angle at the base. Find the angles of the triangle. 2. The angles of a triangle are denoted by 3#, 7x, and 5x. Find each angle. 3. Two right triangles are congruent if a side and the oppo- site angle of the one are equal respectively to a side and the opposite angle of the other. 4. Perpendiculars to the sides drawn from any point in the base of an isosceles triangle make equal angles with the base. 5. Construct a protractor of wood or thick cardboard having a six to nine inch radius. To do this, place the small pro- tractor upon the material to be used and ex- tend the radii to the required length. Fasten a pendulum or marker at the center. This can be used to take vertical angles. RECTILINEAR FIGURES (5. The acute angles of a right triangle have the ratio 2:5. Find them. 7. The angles of a triangle are x, y, s, with x + y = 80 and z y 50. Find x, y, z. 8. To find the height of a flag B staff, measure back from the base of the staff on level ground a convenient distance as AC. From C sight along the straight edge of the protractor to B. The marker swinging freely will indi- cate upon the protractor the angle at B. (Why?) Then find ZC (How?) and on some convenient place construct a rt. triangle hav- ing the parts B, C, AC, and measure the side homologous to AB. 9. To find the height of a tower when the point directly under the top of the tower cannot be obtained. Measure Z m at eome convenient point on level ground and then find Z n. Meas- ure a distance CD, the line BCD being straight, and at J> deter- mine the angle. On convenient ground construct a triangle with the given parts Z n, L D, and side CD. Extend the side DC until it meets the perpendicular dropped to it from the point correspond- ing to A. The length of this perpendicu- lar is the height of the tower. PROPOSITION XXIII. 112. THEOREM. Any side of a triangle is less than the sum of the two other sides and greater than their difference. Given A ABC. To Prove AB < BC + CA and AB>BC~ CA. SUG. 1. ABBC. Why? 1. 2. 3. '.AB>BC CA. 49 (4). Therefore 1. Two sides of a triangle are 6 ft. and 8 ft. respectively. What are the limits to the length of the third side? 46 PLANE GEOMETRY PROPOSITION XXIV. 113. THEOREM. // two angles of a triangle are unequal, the sides opposite these angles are un- equal and the side opposite the greater angle is the greater. Given A ABC with L C> L B. To Prove AB > AC. SUG. 1. Draw CM from C to AB so that L BCM = L B. Show that this is possible. 2. M must fall between A and B. Why? 3. Compare CM + MA with AC. Auth. 4. Compare CM and BM. Auth. 5. Complete the proof. Therefore 114. COB. 1. The hypotenuse is the longest side of a right triangle. 115. COR. 2. The perpendicular from a point to a line is shorter than any other line-segment drawn from that point to the same line. 116. DISTANCE. The length of the perpendicular from a point to a line is the distance from the point to the line. 117. ALTITUDE OF A TRIANGLE. The distance from the vertex of a triangle to the base (produced if neces- sary) is the altitude of the triangle. The word "alti- tude ' ' is often used to indicate the line itself as well as RECTILINEAR FIGURES 47 its length. Since any one of the sides of a triangle may be considered as the base, every triangle has any one of three possible altitudes. 118. MEDIAN OF A TRIANGLE. The sect connecting the vertex of a triangle with the mid-point of the opposite side is a median of the triangle. There are in every tri- angle three medians. 1. The bisectors of the angles at the base of an isosceles tri- angle together with the base form another isosceles triangle. 2. If a line is drawn from any point in the bisector of an angle parallel to one side of the angle and is extended to meet the other side, an isosceles triangle is formed. 3. A ABC is a right triangle with the vertex of the right angle at C. Draw CD to AB so that L ACT) = L A. Prove that D is the mid-point of AB. SUG. Prove AD - CD - BD. 4. Prop. XXII is the converse of what proposition? 5. What relation does the mid-point of the hypothenuse of a right triangle sustain to the three vertices of the triangle? See Ex. 3. 6. To measure a distance between two points, A and B, both of which are inaccessible. A SUG. 1. Lay off a convenient line CD and measure CD, Lo, L 7>i, and Z n. 2. On CD as base construct &ACD and A BCD. 3. This fixes two points corresponding to A and B, giving the required distance. It is to be noted that in these problems the method requires the construction of triangles which, if the distances are large, may be impracticable. 7. Prove that AC > BC AB in figure 112. Why wns the proof as given sufficient to cover this problem? 8. Using the dividers only show that BC < AB + AC. 9. Can a triangle be constructed with sides of 7 ft., 3 ft., and 13 ft.? Of 7 ft., 5 ft., and 12 ft.? Of 3 ft., 8 ft., and 10 ft.? Use the dividers in the construction. 48 PLANE GEOMETRY PROPOSITION XXV. 119. THEOREM. // two sides of a triangle are unequal, the angles opposite these sides are un~ eaual and the angle opposite the greater side is the greater. Given A ABC with AB > AC. To Prove Z C> Z B. Proof. If Z C is not greater than Z B then Z C=/.B or L C Z. Z ACS > /.ACM. .'.ACB>/.B. Why? B^ ^ c PROPOSITION XXVI. 120. THEOREM. // sects are drawn from, the same point in a perpendicular to a line and meet RECTILINEAR FIGURES 49 the line at unequal distances from the foot of the perpendicular, the sects are unequal and the more remote is the greater. M with Given OD 1 EB ED > DP. To Prove OE > OF. SUG. 1. Lay off tween E and D. Why? 2. Draw OM. 3. D any -B point in OD and M will lie be- Then OM = OF. Why ? L 1 is a rt. angle. ' L 2 is acute. Why? 4. 5. 6. 7. 8. Why ? Z 4 is acute. Why? L3>L4. Why? OE>OM. Why? 0# > 0^. Why ? Therefore PROPOSITION XXVII. 121. THEOREM. The shortest line from a point to a straight line is the perpendicular from the point to the line. Let PA represent the shortest line from P to line AB. To Prove PA 1 AB, 50 PLANE GEOMETRY Proof. If PA is not perpendicular to AB then some other line through P is perpendicular to AB. But by 115 this new line would be shorter than PA, hence contrary to hypothesis. Therefore PROPOSITION XXVIII. 122. THEOREM. Two unequal line segments drawn from the same point in the perpendicular to a given line meet the line at unequal distances from the foot of the perpendicular, the longer seg- ment meeting it at the greater distance. V JB D Given AB 1 MN and AC > AD. To Prove BC > BD. SUG. 1. If CB is not greater than BD then CB = BD or else CB < BD. 2. Suppose CB = BD ) how does AC compare with AD? Why? 3. Suppose CB L A'. To Prove BC > B'C'. SUG. 1. Place AA'B'C' upon A ABC so that A'B' coincides with AB. Then since L A > LA' side A'C' will lie between AB and AC. The point C' may fall within A ABC, on line BC, or beyond line BC. The student should construct the figures for the first and second cases lettering them like the first figure. 2. Bisect L CAC f and extend the bi- sector to meet BC at 0. 3. Compare A AOC with A AOC' and OC with OC'. Auth's. 4. Compare EC' with BO + OC' . 5. Compare BO + OC' with BO + OC or its equal BC. Therefore 52 PLANE GEOMETRY PROPOSITION XXX. 124. THEOKEM. // two triangles have two sides of the one equal respectively to two sides of the other and the third sides unequal, the included angles are unequal and that angle is the greater which is opposite the greater side. Given A ABC and AA'B'C', with AB = A'B', AC = A'C', and BOB'C'. To Prove L A > L A'. SUG. 1. What three possible relations are there for A. A and A' ? 2. Test each of these as was done in Prop. XXV. 3. Which is the only one not contra- dicted by the hypothesis? Therefore 125. QUADRILATERAL. A portion of a plane bounded by four sects is a plane quadrilateral. 126. QUADRILATERALS CLASSIFIED. A quadrilateral is a parallelogram if its opposite sides are parallel, a trapezoid if but one pair of opposite sides is parallel, a trapezium if no two sides are parallel. A trapezoid the non-parallel sides of which are equal is isosceles, 127. PARALLELOGRAMS CLASSIFIED. A parallelogram is a rectangle if all its angles are right angles, a rhom- boid if all its angles are oblique, a square if it is an equi- lateral rectangle, a rhombus if it is an equilateral rhomboid. With dividers (or scale) and protractor determine the character of each of the following figures. RECTILINEAR FIGURES 53 128. DIAGONAL. A line-segment connecting any two non-adjacent vertices of a figure is a diagonal. 129. BASE OF A QUADRILATERAL. The side upon which a quadrilateral is assumed to stand is its base. In case of the trapezoid or parallelogram, one of the parallel sides is always considered as the base. The opposite side is the upper base. 130. ALTITUDE. The perpendicular distance between the bases of a parallelogram or a trapezoid is its alti- tude. The word "altitude" is often used to indicate the line itself as well as its length. 131. DIAMETER. The sect which connects the mid- points of two opposite sides of a parallelogram or the mid points of the non-parallel sides of a trapezoid is a diameter. Construct each of the figures above defined and .draw for each the altitude and indicate the bases and diameters. PKOPOSITION XXXI. 132. THEOREM. The opposite sides of a par- allelogram are equal. c Given U ABCD. To Prove AB ~ DC and AD = CB. SUG. 1. Draw a diagonal, as AB, and com- pare A ABC with A CDA. Auth. 2. Compare AD with CB AB with CD. Therefore 54 PLANE GEOMETRY 133. COR. 1. A diagonal divides a parallelogram into two congruent triangles. 134. COB. 2. Segments of parallel lines intercepted betiveen parallel lines are equal. 135. COR. 3. Perpendicular segments between par- allel lines are equal. 136. COR. 4. The perpendicular segment inter- cepted between the base of a triangle and a line through the vertex parallel to the base is equal to the perpen- dicular from the vertex to the base. 137. DISTANCE. The length of the perpendicular segment intercepted by two parallel lines is the distance between them. PROPOSITION XXXII. 138. THEOREM. The opposite angles of a par- allelogram are equal and any two consecutive an- gles are supplementary. See 97 and 102. PROPOSITION XXXIII. 139. THEOREM. The diagonals of a parallelo- gram bisect each other. The construction of the figure is left to the student. Given O ABC D with diagonals AC and BD inter- secting at 0. To Prove AO = OC and BO = OD. SUG. Select appropriate A and compare them. Therefore PROPOSITION XXXIV. 140. THEOREM. // a quadrilateral has two of its opposite sides equal and parallel, it is a par- allelogram. RECTILINEAR FIGURES 55 The construction of the figure is left to the student. Given a quadrilateral ABCD, with AB = DC and AB II DC. To Prove ABCD a /Z7. SUG. 1. What is a parallelogram? 2. How much of the definition is given in the hypothesis and what remains to be proved ? 3. Draw a diagonal as AC, compare L BAG with L ACD and then compare A ABC with A ACD. Complete the demonstration. Therefore 141. A careful distinction must be drawn between statements which are definitions, i.e., which are neces- sary and at the same time sufficient to characterize the thing defined as distinct from all other things, and statements which point out certain properties of the thing in question without being in themselves complete enough to make this distinction. For example since the hypothesis of Prop. XXXIV is sufficient to charac- terize the quadrilateral as a parallelogram according to the definition in 126, it might be taken as the defini- tion while Props. XXXI and XXXII could not be so used for the properties there indicated are possessed by other figures than parallelograms. 1. If one angle of a parallelogram is a right angle, the figure is a rectangle. l\ The sum of the angles of a quadrilateral equals four right angles. 3. If the diagonals .of a quadrilateral bisect each other, the figure is a parallelogram. This bears what relation to 139? 4. If the opposite angles of a quadrilateral are equal the figure is a parallelogram. Sue. Compare the sum of two consecutive angles with four right angles. 56 PLANE GEOMETRY PROPOSITION XXXV. 142. THEOREM. A quadrilateral the opposite sides of which are equal is a parallelogram. The construction of the figure is left to the student. Given O ABCD with AB = DC and AD = BC. To Prove ABCD a O. SUG. 1. Draw the diagonal AC. 2. The conditions of 126 or 140 may be used to identify the parallelogram. 3. What part of either is in the hy- pothesis of this theorem ? 4. What remains to be proved? Com- plete the demonstration. Therefore In the light of 141 what relation does the statement of this theorem bear to the parallelogram? PROPOSITION XXXVI. 143. THEOREM. Two parallelograms which have two sides and the included angle of the one equal to two sides and' the included angle of the other respectively are congruent. D C D' Given 17 AC and A'C', with AB = A'B', A'D',and L A= /. A'. To Prove O AC = O A'C'. SUG. 1. Place OAO upon OA'G" so that AB coincides with A'B'. RECTILINEAR FIGURES 57 2. What direction does AD take? Why? Where does D fall? Why? 3. What direction does DC take and why ? 4. Where does point C fall and why? 5. Complete the demonstration. Therefore PROPOSITION XXXVII. 144. THEOREM. The diagonals of the rhombus and of the square bisect the angles. The proof is left to the student. PROPOSITION XXXVIII. 145. THEOREM. The diagonals of the rhombus and of the square are perpendicular to each other. Given rhombus ABCD (or a square) with diagonals AC and BD. To Prove AC LED. SUG. 1. In A ADD and A AOB compare L AOD and AOB. 2. Complete the demonstration. 3. Note that no reference is made to the character of the angles A, B, C, D. What con- clusion can be drawn as to the application of the demonstration to the square as well ^s the rhom- bus? Therefore 58 PLANE GEOMETRY PROPOSITION XXXIX. 146. THEOREM. A diameter of a parallelo- gram divides it into two congruent parallelo- grams. The proof is left to the student. 147. COR. 1. A diameter of a parallelogram is parallel to the corresponding base. 148. COR. 2. The diameters of a parallelogram bi- sect each other. 149. COR. 3. A line which is parallel to the base of a parallelogram and which bisects one side bisects the other side also. PROPOSITION XL. 150. THEOREM. A line which bisects one side of a triangle and is parallel to the base bisects the other side and equals half the base. Given A ABC with MN I! CB and M mid-point of AC. To Prove AN = NB and 2 MN = CB. SUG. 1. Draw MD \\ AB. 2. Compare A AMN and MCD ; MD and AN ; MD and NB ; AN and NB. Auth. 3. Compare MN and CD; MN and DB; MN and CB. Auth. Therefore ]. The angle formed by the bisectors of the angles at the base of an isosceles triangle is equal to an exterior angle at the of the triangle. RECTILINEAR FIGrRES 50 151. COR. The line connecting the mid-points of the sides of a triangle is parallel to the base and equal to one half the base. SUG. 1. Work out an indirect demonstration. 2. Work ' out a direct demonstration from the above figure, noting that MN II CB if MNBD is a O or if Z AMN L MCD. 150. PKOPOSITION XLI. 152. THEOREM. // a line bisects one of the non-parallel sides of a trapezoid and is parallel to the base, it bisects the other side also f and equals half the sum of the bases. SUG. Draw a diagonal of the trapezoid. The proof is left to the student. 153. COR. The diameter of a trapezoid is parallel to the bases and equal to half their sum. SUG. 1. Through one end of the diameter draw a line parallel to the bases and use the in- direct method. Or 2. Through one end of the diameter draw a line parallel to the opposite side and use the direct method. 1. If the adjacent sides of a parallelogram are not equal the diagonals are not perpendicular to each other. 2. If the diagonals of a parallelogram intersect at right angles the figure is a rhombus or a square. 3. If the diagonals of a parallelogram are equal the figure is a rectangle. 4. The opposite sides of a window frame are equal. Be- fore putting on stay laths the carpenter " squares it." At how many corners should he test it? 60 PLANE GEOMETRY PROPOSITION XLII. 154. THEOKEM. // parallel lines intercept equal segments on one transversal, they intercept equal segments on all transversals. \A \. I \ / : v L Given the parallel lines a, &, c, c?, etc., with transver- sals e and e' such that AB = BC = CD. To Prove A'' = 5'G" C'D'. SUG. 1. e and e' may or may not be paral- lel. What figures are formed in each case? 2. In case e and e' are not parallel draw lines through A, B, C, etc., parallel to e' and terminating on the next following of the parallels. 3. Complete the proof. Therefore 1. Having laid out a plat or four sided frame so that the opposite sides are equal, how can one tell whether or not it is a rectangle without applying a " square" or right angle to one of the angles? 2. A draftsman wishing to draw parallel lines uses a T-square. Explain the principle of its use. How may parallel lines be drawn tvith a carpenter's square? 3. Can a rectangle and a rhombus have sides respectively equal? Construct a rectangle of wood, one nail at each vertex. Can it be distorted into a rhombus? RECTILINEAR FIGURES 61 4. Explain the principle of the parallel rulers. 3. By experiment test which has the greater enclosed area, a rectangle or a rhomboid, the sides of the two being respectively equal. Squared paper can be used. 6. Construct a parallelogram having given two adjacent sides and the included angle, using dividers and straight edge. How could it be done with a protractor and straight edge! 7. Draw a trapezium connecting the mid-points of its adja- cent sides. What kind of a quadrilateral is formed? Sue. Draw a diagonal of the given figure. 8. Prove that the sects that connect the mid-points of opposite sides of a trapezium bisect each other. The following problems are intended for use with pupils who have had or who are taking elementary Physics. 9. Two balls of the same size and rolling with the same speed at right angles to each other on a level surface strike another ball which is at rest. What direction will the latter ball take? If one ball has twice the force of the other, what di- rection will the third ball t;ike? NOTE. The force. which would give to the third ball the same motion as is given to it by the two balls is called the resujtant force or resultant of the forces due to the two balls independently. 10. Draw a parallelogram to represent the forces and di- rections of the three balls in the first part of Ex. 9. This parallelogram is called the parallelogram of forces. 31. Draw the parallelogram of forces for the second part of problem 9, representing a foot by % inch. 12. Suppose the balls in Ex. 9 with equal force strike the third ball at an angle of 45. Draw the parallelogram of forces and a line representing the resultant force. 13. Two forces, one of 5 Ibs. and one of 8 Ibs., strike a body at an angle of 60. Draw the resultant. If they meet at an angle of 100, draw the resultant. If one is a force of 7 Ibs. and the other 21 Ibs. and they meet at an angle of 100, draw the resultant. 62 PLANE GEOMETRY 155. POLYGONS CLASSIFIED. Beginning with the tri- angle and proceeding in order to polygons of four sides, five sides, etc., the plane polygons are the trian- gle, quadrilateral, pentagon, hexagon, heptagon, octo- gon f etc. A polygon of n sides is called an n-gon. 156. A polygon is equilateral if all its sides are equal, equiangular if all its angles are equal, and regu- lar if it is both equilateral and equiangular. Two polygons are mutually equilateral if the sides of the one are equal respectively to the sides of the other, mutually equiangular if the angles of one are equal re- spectively to the angles of the other. The student should prove that two figures both mutually equilateral and mutually equiangular are con- gruent. 157. A polygon is convex if each of its angles is less than a straight angle and concave if one or more of its angles are reflex. In this hook the term polygon will mean a convex polygon. EQUIANGULAR EQUILATERAL REGULAR CONCAVE 158. COR. In any polygon the number of vertices is the same as the number of sides. 1. If ABC is a right triangle with the right angle at C and CD is drawn to the hypotenuse so that LACD LE, then CD 1 AB. Prove L DCB = LA. 2. If two parallel straight lines are cut by a transversal, the bisectors of the interior angles form a rectangle. 3. What relative positions of parallels and transversals will form a square? RECTILINEAR FIGURES 63 PROPOSITION XLIII. 159. THEOREM. The sum of the angles of a polygon is equal to (n 2) straight angles, where n represents the number of sides. Given polygon ABCD . . . , having n sides. To Prove Z^ + Z5+ZC f + etc. = (M - 2) straight angles. SUG. 1. Draw all possible diagonals from some one vertex as A. 2. Then (n 2). triangles are formed. Explain why. 3. How many straight angles in the sum of all the interior angles in all these trian- gles? 4. Complete the proof. Therefore 160. COR. I. The sum of the angles of a polygon is equal to (2n 4) rt. angles, or 2n rt. L 4rt. A. 161. COR. II. Each angle of an equiangular poly- 7 4 2-n - 4 , gon is equal to - rt. A. n 1. What is the sum of the angles of a quadrilateral? A pentagon? A heptagon? A decagon? 2. What is the size of each angle of a regular hexagon? A regular octagon? A regular decagon? 64 PLANE GEOMETRY 3. How many sides has a regular polygon if each angle is 156 ? 4. Four of the angles of a pentagon are 100, 170, 95, and 115 respectively. Find the fifth angle. SUG. Form an equation from Prop. XLIII and solve algebraically. PROPOSITION XLIV. 162. THEOREM. // one exterior angle be formed at each vertex of a polygon, the sum of these exterior angles is equal to four right angles. c Given polygon MC, having n sides with one exterior angle formed at each vertex, as La' formed at ver- tex A with interior Z a. To Prove L a' + L b' + Z c' + . . . = 4 rt. A. SUG. 1. The polygon has n vertices. What is the sum of the exterior and the interior angle at each vertex? 2. What is the sum for all vertices? 3. What is the sum of the interior an- gles? 4. Complete the proof. Therefore 163. COB. The sum of all the exterior angles of any polygon is equal to eight straight angles. 1. How many degrees in each exterior angle of a regular pentagon? Of a regular octogon? 2. How many sides has the polygon, the sum of the exterior angles of which is equal to the sum of the interior angles .' 3. An exterior angle of a regular polygon is 60. How many sides has the polygon? 45? RECTILINEAR FIGURES 65 4j An exterior angle of a regular polygon is of the adjacent angle. Find the number of sides of the polygon. 5., Each angle of a regular polygon is of a straight angle. Find the number of the sides of the polygon. 164. Locus. To locate a point definitely in a plane two conditions limiting its position must be given. If but one condition is given the point is limited to one or more lines. For example, if all that is known of a point is that it is four inches from a given straight line, the point is not definitely fixed in position but lies on either of two lines and may be any point on either. The lines in question are parallel to the given line, lying the one on one side and the other on the other side of the given line, at four inches distance, as will later be established. The locus of a point is defined as the line or group of lines to which a point is limited, any point of which satisfies the given conditions. The locus of a point is both inclusive and exclusive.. It includes all the points which satisfy the given condition and excludes all points which do not. 165. Locus GENERATED. The locus of a point may be regarded as the path of a point which moves according to a given law or condition. For example, the law in the illustration used in the preceding section is that the point must keep at the distance of four inches from the given line. The expressions, locus of a point and locus of points are both used. Discuss without formal proof the following loci: 1. What is the locus of a ship at sea 20 N. Latitude f 2. A man lives one block west of the north and south street which passes the school house. What is the locus of his house? 3 A man is ^ mile from the street or road in front of the school. What is his locus? 66 PLANE GEOMETRY 4. An article was lost ten feet west of the outer edge of a straight side walk. Where should one look for it, i. e. what is its locus? 5. In Ex. 4 change the words "west of" to "from." What is the locus? 6. What is the locus of the center of the headlight of a locomotive when in motion? 7. A farm is in range seven, what is its locus? In township fifteen, what is its locus? 8. What is the locus of the tip of the pendulum of a clock? 9. What is the locus of the hub of a wheel on a moving auto? 166. To VERIFY Locus CONDITIONS. To prove that a given line (or lines) is the locus of points satisfying a given condition it is necessary to prove two things; first any point in the line satisfies the given condition and second no point outside the line satisfies the given con- dition. For an example, see 167. PROPOSITION XLV. 167. THEOREM. The locus of points equally distant from the extremities of a given line seg- ment is the perpendicular bisector of the line segment. M x N Given line segment AB, with MN the perpendicular bisector of AB. To Prove MN the locus of points equidistant from A and B. SUG. 1. See Prop. V. 2. Let X' be a point not on MN. Com- pare X'A and X'B. Auth. 3. Apply definition of locus. Therefore RECTILINEAR FIGURES 67 168. COR. The locus of points equidistant from two given points is the perpendicular bisector of the line joining them. PROPOSITION XL VI. 169. THEOREM. The locus of points equidis- tant from the sides of an angle is the bisector of the angle. See examples 1 and 2 on p. 37. 170. COR. The locus of points equidistant from two intersecting lines is the pair of lines that bisect the an- gles formed by the given lines. What relation do these locus lines bear to each other ? PROPOSITION XL VII. 171. THEOREM. The locus of points at a given distance from a given line is the pair of lines par- allel to the given line and at the given distance from it. SUG. 1. Through any point in the given line draw a perpendicular, on this line locate the two points which are at the required distance from the given line and through these points draw lines parallel to the given line. 2. What two facts must be established in order to prove these two constructed lines to be the desired locus? Therefore 1. If the opposite angles of a parallelogram are bisected by the diagonals, the figure is equilateral. Sue. Draw one diagonal. 68 PLANE GEOMETRY i PROPOSITION XL VIII. 172. THEOREM. The locus of points equally distant from two parallel lines is the line parallel to them and midway between them. SUG. 1. Construct a segment perpendicular to and intercepted by them. 2. Construct the perpendicular bi- sector of this segment. 3. Complete the proof. Therefore 173. USE OF Loci. A common method of locating a point in a plane is to establish two loci of the point, the intersection of which is the required point. To use this method of attack successfully the student must bear in mind that two statements are made when a line, as AB, is said to be the locus of an unknown point X, viz : any point in AB can be X and no point outside of AB can be X, i.e. X must, be in AB. 1. Find point X if it is to be in a given line and equally distant from two points. GIVEN line CD and the points A and B (which may lie, one or both, on CD). To find point X in CD and equally distant from A and B. SUG. 1. What is the locus of points equally distant from A and 5? 2. Where then must X lie? 3. What position of the points A and B with respect to CD would make the problem impossible? 2. Find point X if it is in a given line and equally distant from two intersecting lines. SUG. 1. What is the locus of points equally distant from two intersecting lines? 2. Complete the demonstration. 3. Is this problem ever impossible? What is the condition? RECTILINEAR FIGURES 69 4. What relative position of the given lines would admit of an unlimited number of positions for point XI 3. Find point X if it is equally distant from two intersecting lines and equally distant from the extremities of a given line- segment. Is there any arrangement of the given lines which will make this problem impossible? 4. Find point X if equally distant from the sides of a given angle and equally distant from two parallel lines. 5. Find point X if equally distant from two points and equally distant from two parallel lines. 6. Find point X if it lies in one given line and is equally distant from another given line. Are there conditions which make the problem impossible? 7. Find X if it is equally distant from two intersecting lines and also a given distance from a third line. In general, how many such points are there! Whatsis the least possible number? What is the greatest possible number f 174. CONCURRENT LINES. Two or more lines having one point in common are concurrent lines. The construction of loci is a useful method of proving certain lines concurrent. Name the concurrent lines in the above exercises. 1. The perpendicular bisectors of the sides of a triangle are concurrent. SUG. 1. Two of the bisectors as DE and FG musF mfersect as~aT~#T W-hy ? What loci are here involved? 2. OA = OC and OC = OB. Why? C^ F % B 3. Therefore is equidistant from B and A. Why? 4. In what third line must then lie? Why? 2. The bisectors of the angles of a triangle are concurrent. SUG. 1. Two of the bisectors must intersect. Why? What loci are involved? 2. Prove their intersection to be in the third bisector j 175. SUMMARY OF BOOK I. State all the authorities by which (1) two triangles may be proved congruent. 70 PLANE GEOMETRY (2) two line segments may be proved equal. (3) two lines may be proved parallel. (4) two lines may be proved perpendicular to each other. (5) two angles may be proved unequal. (6) two line segments may be proved unequal. (7) a quadrilateral may be proved a parallel- ogram. (8) a locus has been established. These authorities should be carefully arranged, writ- ten out in a note book, and memorized. They are the "tools" to be used in future constructions and demon- strations. Geometry is not half learned if the student is unable to recall the various hypotheses which are available for deducing any desired conclusion. 1. BD bisects L ABC and EF LBD at B. Prove L FBA = L CBE. Prove the theorem when EF cuts BD at some point other than B. 2. If two straight line segments bisect each other at right angles any point in either is equidistant from the extremities of the other. 3. By how much does the supplement of an acute angle exceed the complement of the same angle? 4. If one of two supplementary adjacent angles is bisected, the perpendicular to this bisector at the vertex bisects the other angle. 5. If the bisectors of two adjacent angles are perpendicular to each other the angles are supplementary. Compare with Ex. 4. 6. The perimeter of a triangle is leis than twice the sum of the medians. 7. The bisector of an angle of a triangle and the bisectors of the exterior angles of the two other vertices meet in a point which is equidistant from the sides of the triangle. Use loci. 8. If a line intersects the sides of an isosceles triangle at equal distances from the vertex ; it is parallel to the base. RECTILINEAR FIGURES 71 9. The line joining the feet of the perpendiculars from the extremities of the base of an isosceles triangle to the sides is parallel to the base. 10. Two angles having their sides respectively parallel, one pair of parallel sides having the same direction and the other pair having opposite directions, are supplementary. 11. Two triangles having two angles and a side opposite one of them in the first equal respectively to two angles and a corre- sponding side in the other are congruent. 12. Two isosceles triangles having equal bases and equal vertex angles are congruent. 13. Two triangles having two sides and an angle opposite one of them in the one triangle equal to two sides and an angle opposite one of them respectively in the other triangle may or may not be congruent. A D GIVEN A ABC and A DEF or D'E'F', with AB=DF = D'F', AC = DE = D'E', L B = L F = L F'. To PROVE A ABC = A DEF but not congruent to A D'E'F'. It is to be noted that of the two triangles DEF and D'E'F' one is acute and one is obtuse, and each have the required parts. Obviously A ABC is congruent to but one. The student should give detailed proof for the figures as drawn. 14. The sum of the exterior angles at the base of a triangle is equal to two right angles plus the vertex angle. 15. In &ABC, LC is twice the sum of L A and ZB and LE is twice LA. Find A, B, C. 16. The exterior angles formed by producing the sides of an isosceles triangle beyond the base are equal. 17. Given a square ABCD. Draw the diagonal CD and on CD lay of CE =?CB. Draw EF LCD at E and extend to DB as at F. Prove DE = EF = BF. 18. A straight line segment intercepted between parallels is bisected and another straight line is drawn through the point of bisection. Prove that the segment of this line intercepted between the parallels is bisected. 72 PLANE GEOMETRY 19. Find the sum of the interior angles of a concave polygon of 13 sides. 20. A segment is intercepted between two parallels and the adjacent angles formed by the segjnent and one of the parallels are 'bisected. Prove that the seg- ments on the other parallel are equal. 21. ABC is an isosceles triangle. DE and DF are parallel to AC and AB re- spectively. Prove the perimeter of AFDE equal to AB + AC. 22. If the legs of a trapezoid are equal, they make equal angles with the parallel sides. 23. Prove the converse of Ex. 22. 24. Prove that the sum of the angles at the vertices of the conventional five pointed star is equal to two right angles. Use the figure as drawn. 25. Connect the vertices of the star and give a different demonstration. 26. How many right angles in the sum of the vertex angles of a six pointed star? Of an eight pointed star? 27. If the vertex angle of an isosceles triangle is one-half as great as an angle at the base, the bisector of a base angle di- vides the given triangle into two isosceles triangles. 28. How many equiangular triangles can have a common vertex? How many rectangles? How many equiangular pentagons? Hexagons? Equiangular figures of a still greater number of sides? 29. Which of the above figures will exactly fill up the space about the common vertex? 30. Which of the above figures can be used for a patch work quilt or a mosaic design? CHAPTER II. CIRCLES. 176. CIRCLE. A closed curve in a plane all points of which are the same distance from a fixed point in the plane is a circle. The fixed point is the center of the circle. The circle completely encloses a portion of the plane. Thus a circle is the locus of points in a plane at a given distance from a fixed point in the plane. It has been more or less customary for writers on geometry to define the circle as the portion of the plane enclosed by the curve and to call the curve the circum- ference of the circle. The above use of the term is more convenient and accords more perfectly with its general use. 177. LENGTH OP A CIRCLE. The length of the curve is called the length of the circle. 178. RADIUS. A straight sect joining the center and any point of the circle is a radius. 179. CHORD. A straight sect terminated at both ends by the circle is a chord. 180. DIAMETER. A chord which passes through the center is a diameter. 181. ARC. Any portion of a circle is an arc. An arc which is half of a circle is a semicircle. An arc less than a semicircle is a minor arc and one which is greater than a semicircle is a major arc. 182. CENTRAL ANGLE. An angle the vertex of which 74 PLANE GEOMETRY is the center of a circle and the sides of which are radii is a central angle. 183. SECANT. A straight line that intersects a cir- cle twice is a secant. A secant is a chord produced. .x v OA and OB are radii, AB is a / o__-V diameter, CD is a chord and ED is A T cy a secant. ^^_ ^^ 184. SUBTENDS. A chord subtends the two arcs which have the same extremities as itself. Unless other- wise stated the arc subtended by a chord is understood to be the minor arc. 185. PRELIMINARY THEOREMS AND ASSUMPTIONS. 1. All radii of a circle are equal. 2. The diameter of a circle is twice the radius. 3. All diameters of a circle are equal. 4. The distance from any point in the plane to the center of a circle is greater than, less than, or equal to a radius according as the point is outside the circle, on the circle, or within the circle. 5. If the radii or diameters of two circles are equal the circles are equal. Since all circles are of the same shape this implies congruence as well. 6. If two circles are equal, the radii and the diam- eters of the two are equal. 7. An unlimited straight line that lies partly within a circle cuts the circle in two points. 8. If the end points of two minor arcs or two major arcs on the same or on equal circles can be made to coin- cide the arcs can be made to coincide. 9. If two arcs of two circles are equal the circles are equal. CIRCLES 75 186. POSTULATE. A circle may be constructed about any given point in the plane as center ivith any given sect as radius. The actual construction of such a circle is done by aid of the dividers. 187. COR. A circle may be constructed upon any given sect as diameter. SUG. What must be done to the given diameter to find the center of the required circle ? 188. PROBLEM. Given an arc of a circle, to find the center. SUG. If from the given arc MC two loci of the center can be determined, the center can be M found. 2. Draw two chords as AB and AC. Where must the center lie with respect to A and Bf With respect to A and Cf 3. Construct the loci of points which satisfy these respective conditions. What can be said of the intersection of these locij 4. A circle about this point as a cen- ter with radius OA will pass through points A, B, C and therefore by 185 (8) will embrace the given arc. Therefore, point is the required center. 189. PROBLEM. To draw a circle through three points not in the same straight line. SUG. Make the construction from 188, with A, B, C as given points in required O. 190. COR. I. A circle cannot be drawn through three points which are in a straight line. 76 PLANE GEOMETRY i 191. COB. II. Only one circle can be drawn through three points not in a straight line. SUG. It was seen in 189 that at least one such circle can be drawn. From 185 (8) it fol- lows that there cannot be more than one. Why? Or, take a point X as any other center. Can X be equally distant from A and B f from A and C f 1. How many circles can be drawn through two points? 2. Two circles can intersect in two points at most. 3. Given arc M, find the center. PROPOSITION I. 192. THEOREM. In the same circle, or in equal circles, equal central angles intercept equal arcs. Given circle equal to circle 0' and L AOB = LA'O'E'. To Prove arc AB = arc A'B'. Proof. SUG. 1. Place OO on OO' with on 0'. Then the two circles will coincide, having a com- mon center and equal radii. 2. Rotate OO about 0' until point A falls on A'. 3. Then since L AOB = L A'O'B' the radii OB and O'B' will coincide and B fall on B' . 4. Hence the arcs AB and A'B' coin- cide. 185(8). Therefore 1. How can a carpenter test the accuracy of his "square" without using a previously made right angle? CIRCLES 77 PROPOSITION II. 193. THEOREM. In the same circle, or in equal circles, equal arcs subtend equal central angles. Given O and O 0' with arc AB = arc A'B' '. (Use fig. Prop. I.) To Prove Z AOB = Z A'O'B', SUG. Place the circles together as in Prop. I and show that the two angles can be made to coincide. Therefore PROPOSITION III. 194. THEOREM. //, in the same or in equal circles, two central angles are unequal, the greater angle intercepts the greater arc. Given O C equal to O C" and Z ACB < L A'C'B'. To Prove arc AB < arc A'B'. Proof. SUG. 1. Place O C upon O C' as in Prop. I and rotate it until CA coincide with C'A'. 2. Where will CB fall with reference to ZA'C"'? Why,? 3. Where will B fall with reference to thearcA'5'? Why? 4. Compare arcs A'B and A'B' ; arcs AB and A'B'. Therefore 1. A diameter is greater than any other chord. Sue. Draw any chord not a diameter and draw radii to its extremities. Compare the chord with the sum of the radii. 78 PLANE GEOMETRY PROPOSITION IV. 195. THEOREM. //' two arcs, in the same or in equal circles, are unequal, the greater arc sub- tends the greater central angle. Given O C equal to O 6" and arc AB < arc A'B'. (Use fig. of Prop. III.) To Prove Z ACE < L A'C'B'. Proof. SUG. Place O C upon OC" as in Prop. Ill and show that Z ACS equals a part of /.A'C'B'. Therefore PROPOSITION V. 196. THEOREM. In the same circle, or in equal circles, equal chords subtend equal arcs. Given O F and OF' equal with chord AB = chord A'B'. To Prove arc AB = arc A' ', Proof. SUG. 1. Draw the radii FA, FB, F'A' F'B'. 2. The arcs are equal provided LF-=LF'. Why? 3. Prove these angles equal and complete the demonstration. Therefore 1. Napoleon and his engineer in exploring came to a river. Napoleon asked its width. The engineer sighted from the rim of his cap to the opposite side, swung upon his heel, and sighted to a point on the land, then paced to the point and said "Ten rods, Sire." Upon what proposition did his computation depend? CIRCLES 79 PROPOSITION VI. 197. THEOREM. In the same circle, or in equal circles, chords which subtend equal arcs are equal. Given two equal , F and F' with arc AB = arc A'B' . (Use fig. of Prop. V.) To Prove chord AB = chord A'B'. Proof. SUG. Compare the central angles F and F' and complete the demonstration. Therefore PROPOSITION VII. 198. THEOREM. In the same circle, or in equal circles, two chords which subtend unequal minor arcs are unequal . and the greater chord subtends the greater arc. Given equal E and E' with arc AB < arc A'B'. To Prove chord AB < chord A'B'. Proof. SUG. 1. Draw radii EA, EB, E'A', E'B'. 2. Compare 1 E and E'. Auth.? 3. Compare chords AB and A'B'. Therefore 1. If the mid-points of the three sides of a triangle bo joined by straight lines, the triangle is divided into four congruent tri- angles. 2. If the mid-points of two opposite sides of a quadrilateral be joined to the mid-points of the diagonals, the joining lines form a parallelogram. As one particular case let the quadrilateral be a parallelogram. Ts this case an exception? 80 PLANE GEOMETRY PROPOSITION VIII. 199. THEOREM. In the same circle, or in equal circles, two arcs which subtend unequal chords are unequal and the greater arc subtends the greater chord. Given equal E and E' with chord AB < chord A'B'. (Use fig. of Prop. VII.) To Prove arc AB < arc A'B'. Proof. SUG. 1. Compare A. E and E' . By what theorem ? 2. Complete the demonstration. Therefore PROPOSITION IX. 200. THEOREM. In the same circle, or in equal circles, equal chords are equally distant from the center. Griven equal C and C' with chord AB chord A'B'. To Prove AB and A'B' equally distant from C and C' respectively. Proof. SUG. 1. Draw CM LAB and C'M'LA'B'. Why? Also draw CB and C'B.' Why? 2. It is necessary to prove CM=C'M'. Why? 3. Complete the proof. Therefore CIRCLES 81 PROPOSITION X. 201. THEOREM. In the same circle, or in equal circles, chords which are equally distant from the center are equal. Given C and C' with chords AB and A'R' such that their distances CM and G'M' from the respective centers are equal. (Use fig. of Prop. IX.) To Prove chord AB = chord A'B'. Proof. SUG. Compare MB and M'B'. Complete the demonstration. Therefore PROPOSITION XI. 202. THEOREM. In the same circle, or in equal circles, unequal chords are unequally distant from the center and the shorter chord is at the greater distance. Given O C with chord AB < chord ED, CM and CN be- ing the respective distances of the chords from the center. To Prove CM > CN. ^ : D Proof. SUG. 1. Draw chord FD = AB and CG 1 FD. Connect O and N. 2. Compare Z x with Z y. Auth. 3. Hence Z u < Z v. Why ? 4. '''CG>CN. Why? 5. '. CM>CN. Why? Therefore PLANE GEOMETRY PROPOSITION XII. 203. THEOREM. In the same circle, or in equal circles, chords which are unequally distant from the center are unequal and that chord ivhich is at the greater distance is the shorter. Given O C, CM and CN being the respective distances of two chords AB and ED from the center. Also CN < CM. (Use fig. of Prop. XL) To Prove AB < ED. Proof. SUG. 1. \7hat three possibilities exist as to the relative sizes of AB and EDI 2, Assume each in turn to be true. Which is the only one which does not by a former theorem lead to a conclusion contradictory to the hypothesis? Therefore Prove prop. 203 by direct method (use fig. 202). PROPOSITION XIII. 204. THEOREM. A radius perpendicular to a chord bisects the chord. Given O C with chord AB and radius CN 1 AB at .V. To Prove AN = NB. Proof. SUG. 1. Draw radii CA and CB. Why ? 2. Compare A CNA with A CNB. Auth. 3. Compare AN and NB. Auth. Therefore CIRCLES 83 205. COR. 1. A radius perpendicular to a chord bisects the arc subtended by the chord. SUG. Compare the two arcs by means of the subtended central angles. 206. COR. 2. A radius ivhich bisects a chord is perpendicular to the chord. 207. COR. 3. The perpendicular bisector of a chord of a circle bisects the arc subtended by the chord. 208. COR. 4. The perpendicular bisector of a chord of a circle passes through the center of the circle. 1. If the straight line connecting the mid-points of two chords of a circle passes through the center, the two chords are parallel. 2. The radius drawn to the mid-point of an arc is the per- pendicular bisector cf the subtended chord. 209. TANGENT LINE. A line that touches a circle at one point only and does not cut the circle is a tangent line. 210. POINT OF TANGENCY. The point which is com- mon to the circle and a tangent line is the point of tangency. 211. TANGENT CIRCLE. If two circles have but one point in common they are tangent circles. If one circle is within the other they are tangent internally. If they are with- out each other they are tangent exter- nally. 1. In a right triangle with acute angles of 30 and 60 re- spectively, one side is one-half the hypothenuse. 2. If the hypothenuse of a right triangle is equal to twice one of the sides the acute angles are 30 and 60 respectively. 3. The mid-point of the hypothenuse is equidistant from the three vertices. 84 PLANE GEOMETRY PROPOSITION XIV. 212. THEOREM. The straight sect joining the centers of two tangent circles passes through the point of tangency. Given two circles C and C', tangent at M. To Prove line CC' to pass through M. Proof. CASE I. Internal tangency. SUG. 1. Assume that CC' does not pass through M. Join C to M and C' to M. Extend CC' to meet O C at N and O C' at N' . The pupil will note that the figure is distorted for the sake of the argument. The points C' C' are not the true centers. 2. Compare CM and CN. Auth? 3. CN>CN'. "Why? 4. CN' CC' + C'N'. 5. C'N' = C'M. Why? 6. " <73f > CC' + C'M. Is it possible ? 7. What of the assumption in step 1? CASE II. External tangency. 1. Make the same assumption as in case I. 2. CM = CN and CN < CN'. Why ? 3. CW' + CW = CC'. 4. CN+C'N', or 3. The 2 ft. 2 statement above as to the ratio of the sects a and b will be short- MEASUREMENT 99 248. COMMENSURABLE. Two magnitudes or quanti- ties are commensurable if they each contain a common unit of measure a whole, or integral, number of times. The ratio of two commensurable quantities must then be an integral number or a quotient of two integral numbers. In determining the common unit of measure of two commensurable magnitudes or quantities, the usual method for finding the greatest common divisor may be followed, which is to divide the greater of the two by the less, then the divisor by the remainder, this re- mainder by the second remainder and so on, until a re- mainder zero is obtained. This exact divisor is the de- sired common unit of measure. 249. INCOMMENSURABLE MAGNITUDES. Two magni- tudes or quantities which do not possess a common unit of measure are incommensurable. The ratio of two such magnitudes is neither an integral number nor a frac- tion. The circle and its diameter are incommensurable, as are also the diagonal and side of a square. A proper discussion of this subject cannot be made at this point on account of certain diffi- culties necessarily involved. 250. PROPORTION. A proportion is an equality each member of which is a ratio. The four numbers or quan- tities a, b, c, d are in proportion if the ratio of a to b equals the ratio of c to d. The symbolic form of this statement is - -. The four numbers or quan- b d tities are the terms of the proportion. a. An equality of three or more equal ratios is a con- tinued proportion. 100 PLANE GEOMETRY 1. Arrange the numbers 2, 5, 20, 8 in a proportion in as many ways as possible, verifying each by the definition of proportion. 2. Arrange the numbers 3, 5|, 28^, ] 7| in a proportion and verify it. PROPOSITION I. 251. THEOREM. // a line is parallel to the base of a triangle it divides the sides into propor- tional segments. /-A Given A ADE with line BC II DE. To Prove H = 4<}. BD CE Proof. SUG. 1. To obtain the ratio of the sects AB and BD they must be measured. Let I be the unit sect and suppose it is contained m times in AB and n times in BD. What is the ratio ? Why ? BD 2. Draw lines through the points of division in AB and BD il to DE and extend to AE. Compare the segments on AC and CE. 3. What is the ratio of AC to CE? Why? AB AC 4. Compare the ratio - - with BD CE Therefore NOTE. In the above demonstration the case in which AB and BD have no common unit of measure is not considered. The con- clusions in this case, however, are the same as above but the PROPORTION 101 demonstration is omitted here as being too difficult for the be- ginning student. 252. AN IMPORTANT CONSIDERATION. Since all ratios of magnitudes or quantities, by the definition in use, can be considered only through the ratios of their numerical measures, all proportions herein will be treated as numer- ical proportions. And since the ratio of two magnitudes, as a and b, 247 (b) is a number and can be represented only by the ratio of their numerical measures, as n the terms a and b will be considered, in the interests of brevity, to represent, as well, their numerical measures. In general the names and notations of magnitudes used in the operations of proportion will be synonymous with the notations for their numerical measures, and the treat- ment of the proportions used will be that of algebra. For convenience of reference the propositions of pro- portion will be collected and briefly reviewed in the fol- lowing sections. 1. What is the ratio of one side of an equilateral triangle to the perimeter? Of the perimeter to one side? 2. What is the ratio of a right angle to an angle of an equilateral triangle? 3. W T hat is the ratio of a quadrant to a semi circle? To a circle? 253. THE TERMS OF A PROPORTION. The first and third terms of a proportion are the antecedents. The second and fourth terms are the consequents. The second and third terms are the means. The first and fourth terms are the extremes. In the proportion, - a and c are the ante- b d cedents; b and d are the consequents; b and c are the means; a and d are the extremes. 102 'PLANE GEOMETRY THEOREMS OF PROPORTION. 254. THEOREM I. The product of the means equals the product of the extremes. SUG. Write the ratios as fractions and clear the equality of its fractional form. 255. THEOREM II. // the product of two numbers equals the product of two others, the factors of one product may be made the means and the factors of the other product may be made the extremes of a proportion. Given ab = cd. To Prove * = c b SUG. By what must ab be divided to pro- duce ? By what must cd be divided to pro- c duce - ? Why are the resulting ratios equal ? b 1. From ab = cd derive a proportion in which a and b are the means. 2. From ab = cd make as many proportions as possible. Note in what respects they differ. 3. If the first three terms of a proportion are 5, 7, 15 what is the fourth term? _!__. Find*. 4. Given n - x 5., Find x in the following proportions: JL ..-.*, . -- A 11 = . 2. I:? _ 11.1 _ jl A _ 1? 13 ~ 18' 11 , 13> 15 ~ 27' x ~ 39' x ~ 12' 11 ~ x' 6. Make four different proportions from the identity 8 X 7 = 4 X 14. 7. Use theorem I to determine whether or not the ratios and will form a proportion.1 12 15 256. THEOREM III. // - = - M- - = - b d c d PROPORTION 103 SUG. Use Theorems I and II. 257. ALTERNATION. The interchange of the two means of a proportion is alternation. 1. State theorem lit in words without the use of symbols a, b, c, d. 2. In the fig. of 251 take the proportion by alternation. 0. Construct a triangle with a line parallel to the base. Measure three of the four segments into which the two sides are cut and by 251 determine the fourth. Check the calculation by measurement. 4. Show how the conditions of Ex. 3 may be used to measure an inaccessible distance, over a pond for example. 5. On the school grounds drive four stakes A, B, C, not in a straight line and D in the line AC. Find the point for a 5th stake so as to measure AD indirectly. 258. THEOREM IV. // - = - then - = -- 1) d a c . a I SUG. 1 - b a Complete the proof. 259. INVERSION. The interchange of first and sec- ond, third and fourth terms of a proportion is inversion. 1. State theorem IV in words without the use of the symbols a, b, c, d. 2. In Ex. 5, 255, take the proportions by inversion. Which of those given may be obtained from the others by inversion? 260. THEOREM V. // - = - then = . 1) d b d SUG. Add 1 to each member of the given proportion. Complete the proof. 261. COMPOSITION. The second proportion in Theo- rem V is obtained from the first by composition. This process is sometimes termed addition. 1. State theorem V in words without the use of the symbols a, b, c, d. 104 PLANE GEOMETRY 2. By composition, alternation, etc., prove in the triangle of Prop.I the proportion 4R = EA BD CE 3. Prove AD = AE . Sug. 256 and 260. AB AC 4 . Prove 4 = 42. AC AE BD AB 6. Prove = - AE CE 262. THEOREM VI. // - = -, then b d b d SUG. Subtract 1 from each member of the given proportion. Complete the proof. 263. DIVISION. The second proportion in Theorem VI is obtained from the first by division. This is some- times termed subtraction. 7 , a c ., a+b c+d 264. THEOREM VII. // - = - then b d ab cd SUG. The second proportion is said to be obtained from the first by composition and divi- sion. It may be obtained from the final propor- tions in Theorems V and VI. Complete the proof. In the accompanying figure, CD being parallel to NO, show that a_ _ b. J> _ d, , a+c _ b+d . a+c _ b+d . a _c__ . a & _ cd . c d' a c' a b ' c d ' a+b c+d' a + b c+d' a b _ a+b cd ~ c+d 2. In the proportion L = _ b d M a = 12, b = 5, c = 6, find d. a = 3, c = 7, d = 14, find b. b = 12, c = 13, d = 6, find a. a = 13, b = 15, d - 45, find c. PROPORTION 105 PROPOSITION II. 265. THEOREM. // a line divides two sides of a triangle proportionally, it is parallel to the base. Given _ = _ AD AE To Prove BC II DE. Proof. SUG. 1. If BC is not parallel to DE. let BM represent the line through B which is. 2. Then ^ = ^. Why? AD AE 3. From this proportion and the hy- pothesis compare AC and AM. 4. "Where then must point M lie wi':h respect to Cf Therefore 1. Given Ex. 2, 264. MN = 1 7, MO = 24, DO = 8 ; find a, ~b, c. 2. Given c = 12, tZ = 6, and MN = 30, find the other parts. 266. FOURTH PROPORTIONAL. The fourth term of a proportion of four different terms is a fourth propor- tional to the three others in order. 3. Given three sects a, b, c, find x a fourth proportional by construction. Sug. Construct an angle one side of which is the sum of the sects a and b. On the second side at the ex- tremity of sect a lay off sect c. Complete the construction and verify it. 4. Divide a given sect in the ratio of three to four. Of two to three. Of five to three. 5. Divide a given sect a into two parts having the ratio of the given sects b and c. 106 PLANE GEOMETRY PROPOSITION III. 267. THEOREM. The 'bisector of an angle of a triangle divides the side opposite into sects which are proportional to the adjacent sides. Given A ABC, with AM bisecting Z A, and M the point of division of CB. To Prove - MB AB Proof. SUG. 1. Extend CA to making AO = AB. Join and B. 2. 05 II AM. Why? 3. Note that AM is parallel to BO and derive the required proportion. Therefore 1. If a line bisects one side of a triangle A and is parallel to the base it bisects the other /\ side and equals half the base. B X__\c D// A E Sue. To prove BC \ DE, draw a line through B par- allel to AE. Then use the first part of the exercise. See 150 for a different method. 2. If a line bisects two sides of a triangle, it is parallel to the base and equals one-half the base. See 151. Use another method here. 3. Divide sect a of Ex. 5 P. 105 into tlirco equal parts. Into four equal parts. GIVEN AB = BD and BC II DE. PROVE AC - CE and BC = PROPORTION 107 268. INTERNAL DIVISION. A sect is divided into seg- ments internally when the point of division lies in the segment. AB is divided internally at C into the two segments AC and CB. Thus AC + CB must equal AB. Ax. 9, 49. 269. EXTERNAL DIVISION. A sect is divided into seg- ments externally when the point of division lies in the extension of the sect. A g c AB is divided externally at C into segments AC and CB. In order that in this case also the sum AC + CB may equal AB, the algebraic idea of. positive and negative quantities may be introduced. In moving a point from A to B through the point of division C, the directions AC and CB are the same for internal division and op- posite for external division. In the latter case, in mov- ing from C to B, the point traverses a second time part of segment AC but with respect to the direction of AC, sect CB is then negative. Hence in this case also one may write AC + CB = AB. Unless expressly stated the notations used in this text will not involve this use of directed lines. 1. Draw a tangent to a circle at a given point on th'e circle. 2. Two tangents drawn to a circle from the same point are equal. 3. If two circles are concentric, all chords of the larger which are tangent to the smaller are equal. 4. The tangents to a circle at the extremities of a diameter are parallel. 5. Draw a triangle as large as may be on a given sheet of paper and draw a line parallel to the base and intersecting the sides. Measure three parts from which measurements the fourth part may be found. Check by measuring this fourth part. 108 PLANE GEOMETRY PROPOSITION IV. 270. THEOREM. The bisector of an external angle of a triangle divides the opposite side ex- ternally into sects proportional to the adjacent sides. Given A ABC with L CBE bisected by a line divid- ing the opposite side AC externally in point 0. AO AB To Prove = OC BC Proof. SUG. 1. Draw CM II BO. CO = MB BC Therefore In the above figure and demonstration the inequality AB > BC is assumed. Let the pupil prove the theorem for the case AB L AC by lettering the figure the same and following the line of proof. 1. It is desired to measure an inaccessible distance on a plane. What use can be made of Proposition 270? What measurements shall be taken to determine AC? PROPORTION 100 1. A line drawn through the vertex of a triangle dividing the opposite sides into segments proportional to the adjacent sides bisects the angle. This is the converse of what proposition ? 2. If a radius of one circle is a diameter of another, the circles are tangent and any line drawn from the point of con- tact to the outer circle is bisected by the inner one. 271. HARMONIC DIVISION. A line is divided har- monically when it is divided internally and externally in the same ratio. * M' 1. AB is divided internally in the ratio of 6 to 4 at points. AB = 6 + = 10. 2. AB is divided externally at M' in the ratio of 6 to 4. AB = 6 + (-4) = 6-4 = 2. The unit used in the internal division is contained in AB ten times. In the case of external division the unit is contained in AB two times. In general, for internal division the number of divisions made in AB is the sum of the number made in AM and the number made in MB; for external division the number of divisions made in AB is the difference between the number made in AM ' and the number made in M'B. 3. Divide sect CD harmonically in the ^ D ratio of 5 to 3. Sue. To find M divide CD into 8 equal parts and to find M' divide CD into 2 equal parts. 4. Divide a given sect harmonically in the ratio of 7 to 5. into how many parts must it be divided? 5. If both the interior and exterior angles at a vertex of a triangle are bisected, the opposite side is divided harmonically by the bisectors. A ~" 6. Divide harmonically sect AB in ths ratio of c to d. ~ 110 PLANE GEOMETRY PROPOSITION V. 272. THEOREM. // several lines are drawn parallel to the base of a triangle intersecting the sides, the corresponding segments of the sides form a continued proportion. Given A OMN with lines parallel to the base MN cut- ting the sides into the sects a, b, c, d, e and a', &', c', d', e' , etc , respectively. a b c d c To Prove -=-=-=- = -> etc. a' & c a e Proof. SUG. 1. - = - Why? a b 2. a - = ^. Why? a' a'+b' 3. -^ =~. Why? a'+b' c' 4. Complete the demonstration. Therefore 273. SIMILAR POLYGONS. Polygons which are mu- tually equiangular ( 156) and which have their cor- responding sides proportional are similar polygons. 274. HOMOLOGOUS. In similar polygons those points, lines, and angles which are similarly situated are homol- ogous. In similar triangles the homologous sides are PROPORTION 111 those lying opposite equal angles and the equal angles are those lying opposite homologous sides. 275. ' RATIO OF SIMILITUDE. In similar polygons the ratio of similitude is the ratio of any two homologous sides. r> D' A' 1 The polygons P and P' are similar, provided L A = L A', L B = L B', L C = L C', etc., and AB BC CD AB ' etc. Any one of the equal ratios A'B' B'C' C'D' A'B' etc., may be taken as the ratio of similitude. 276. From the definition of similar polygons, it fol- lows that, if two polygons are known to be similar, the homologous angles are equal and the homologous sides are proportional. 1. In the similar & ABC and A' B'C', if L A = L A', etc., which sides are homologous? If the sides AB and A'B', etc. are homologous, which angles are equal? 277. Polygons may be mutually equiangular but may not have their sides proportional or they may have their sides proportional without being mutually equiangular. The first condition is illustrated by the rectangle and the square. The second condition is illustrated by a rhom- bus and a square or by a rectangle and a rhomboid if the sides are proportional. It will be established later that triangles form an ex- ception to the above statement, in that if either condition of the definition of similarity applies the other is a neces- sary consequence. 112 PLANE GEOMETRY PROPOSITION VI. 278. THEOREM. Two triangles which are mu- tually equiangular are similar. ABC and A'B'C' with Z A = Z A', Z = Z B', Z (7 = Z C'. To Prove A ABC and A'B'C 1 ' similar. Proof. SUG. 1. What part of the definition of sim- ilar triangles remains to be proved? 2. Place & A'B'C' upon A ABC so that A' falls on A, B' on AB at M and 6" on AC at tf. Can this be done ? Why do it ? 3. MN\\BC. Why? 4 . . Why? ,.. Al/ AN A'B' A'C' 5. What is yet to be proved ? 6. Place A A'B'C' upon A ABC with B' on B, etc. Why? 7. What ratios can here be proved equal ? Give all the steps. 1 P 8. Compare the three ratios yl x> AC. _AC B'C'' A'C' Therefore PROPORTION 113 279. COR. I. Two triangles are similar if two angles of one are equal respectively to two angles of the other. 280. COR. II. Two right triangles are similar if an acute angle of one equals an acute angle of the other. 281. COR. III. Two triangles arc similar if they are each similar to the same triangle. Given sect wi. To %,* divide m into segments w ~~..-' proportional to a, b, c, 0^1 Sue. At one extrem- ity O of m draw any line oblique to m and on this line from lay off in order the given sects a, b, c, etc* Why? Complete the construction by reproducing the conditions of proposition V Verify the results. PKOPOSITION VII. 282. PROBLEM. To construct upon a given line a triangle similar to a given triangle. E. G Given A ABC and the line segment EG. To Construct upon EG a triangle similar to A ABC. SUG. Use Prop. XXXV and make the required construction. 1. Draw a tangent to a given circle that shall make a given angle with a given line. 2. Two isosceles triangles are similar if the vertex angles of the two are equal. 3.i Divide a sect into segments proportional to three or more given sects. 4. To measure the height of a nearby object, as EF, lie upon 114 PLANE GEOMETRY the ground in such a position, AC, that the uppe~ end, B, of a pole of known length placed vertically be- F tween the feet will appear in a line with the point F. The distances AC, CB and AE are known or easily measured. How may EF be determined? A- As an illustration of the preceding problem, a woodsman in determining the height of trees uses a pole as long as his own height. What one distance will he need to measure? Why? 5. How may shadows be used to determine the height of objects? 6. Determine as accurately as possible the height of some point on the school building. NOTE. Not more than two should work together, the results being compared in class. D 7. To measure a given height ED. Sue. Set up a pole parallel to ED at some convenient point as C and while one person sights from a point A to D let a second person move a card upon the pole until a point B is found on the pole in the line AD. Make the required measurements and determine the height ED. 8. It is desired to find the distance AB indirectly. If DE is parallel to AB what measurements should be made? 9. A triangle has two angles of 69 and 57 respectively, the included side being 26 rods in length. The length of the two other sides is required. Sue. Construct on paper a triangle similar to the given triangle, using the protractor for the construction of the angles. Measure the three sides. From this data deter- mine the desired distances for the given triangle. State the various methods thus far used for the indirect measurement of distances. Which is the easiest? 10. To construct sects .01, .02, .03, etc., to .09 inches in length Divide a segment one inch long into 10 equal parts and at one PROPORTION 115 extremity erect a perpendicular .1" in length. Connect the other extremities of these two sects forming a right triangle. The per- pendiculars to the original sect at the points of division terminated by the hypotenuse are of the required lengths. Give the reasons for each step. Take upon the dividers .03, .05, .07 of an inch. 11. To construct a diagonal scale ___ T __ r _ T -~rT~TT~|.r by which any segment may be meas- " rT ^ r ~ rT ^l ' ' ' ured in tenths and hundredths of an inch. The construction should be made on cardboard and preserved for future use, if the pupil has not already purchased a diagonal scale. Con- struct a square on a one inch segment, dividing each side into tenths. Connect one vertex of the square with the first point of division on the opposite side, and through the remaining division points of this side draw lines parallel to the first line. Through the division points of the second pair of opposite sides draw parallels to the first pair. Read from the scale .36, .42, .73, .85, .92. Give authority for all statements made. 12. With the dividers take .27 inches from the scale and on some given line lay off a sect .27 inches long: 13. Construct a sect 3.75 inches long; 3.56 inches; 2.05 inches. 14. With the dividers and diagonal scale measure the sects a, &, c, d. a 15. Open a jointed two foot rule so that * the ends are one foot apart. How long is the d sect which connects points one inch from the joint? 2 inches? 5 inches? 9 inches? Use 276. 1G. If the ends of the rule are six inches apart, how far apart are the pair of points marking divisions equally distant from the joint? 17. Open the rule various distances as above and determine the distances between corresponding divisions. IS. Work out the same exercises, using a one foot jointed rule. Also with a six iiK-h jointed rule. 116 PLANE GEOMETRY PEOPOSITION VIII. 283. THEOREM. // tiro triangles have an angle of one equal to an angle of the other and the sides including the equal angles proportional, the triangles are similar. A B C B' C' Given A ABC and A A'B'C' with ZA = A' and AB' AC' To Prove A ABC ^ A A'B'C'. Proof. SUG. 1. What must be proved in addition to the hypothesis to make the triangles similar according to 278 ? 2. Place A A'B'C' upon A ABC so that A' falls on A, A'B' on AB, and A'C' upon AC. Is this possible? Why do so? 3. Where do B' and C' fall? 4. B'C'^BC. Why? 5. Compare Z B' with Z B. Com- plete the demonstration. Therefore 1. If triangles have their sides respectively parallel or per- pendicular to each other they are similar. 2. Let ABC and A'B'C' be two similar triangles. AB-1 ft., A'B' = 14 ft., AC = 5. Find the length of A'C'. If AB is is' ft., AC and A'B' are 11 ft. and 10 ft. respectively; find the length of A'C'. PROPORTION 117 PROPOSITION IX. 284. THEOREM. Two triangles are similar if the corresponding sides are proportional. Given A ABC and A'B'C' with - =-= A'B' B'C' A'C'' To Prove A ABC A 'A'B'C'. Proof. SUG. 1. If any angle of A ABC equals the homologous angle of A A'B'C' the triangles are similar. Why ? 2. Upon AB lay off AM equal to A'B' and upon AC lay off AN equal to A'C'. Connect M and N. A AMN ^ ' A ABC. Why? 3. Compare the ratios - - and - AB BC , A'B' , B'C' also and AB BC 4. Compare MN and #' or c'cos A=b c c and c cos B = a. If cos A = .5 then b =r.5 c. If c = 40 in. then b = i of 40 in. = 20 b 'c A in. 9. One angle of a right triangle is 45 and the hypotenuse is 60 ft. How long is the side adjacent to the given angle? How long is the side opposite? 10. If the adjacent side is 20 in. and the hypotenuse is 63 in. what is the cosine? How many degrees in the angle? 11. If the angle is 33 and the adjacent side is 37 ft. how long is the hypotenuse? How long is the side opposite? 290. THE TANGENT OF AN ACUTE ANGLE. In a right triangle the ratio of the side opposite an acute angle to the side adjacent is the tangent of that angle. TRIGONOMETRIC RELATIONS 123 In the figure of 289 the relations involving tangents are written tan A = - and tan B = or Irtan A = a I) a and a -tan B=b. 1. Construct a right triangle with an angle of 30. Meas- ure the two sides and compute the tangent of 30. 2. What is the height of a flag staff if the angle subtended by the staff at a distance of 75 ft. is 35? 3. In a right triangle the sides are 3 in. and 4 in. respect- ively. What are the tangents of the acute angles? 4. A tower 75 ft. high stands beside a river. The line from the top to a point on the opposite bank makes an angle of 5i )0 with the tower. How wide is the river? 291. ANGLE OF ELEVATION. If at any point in a horizontal or level line a line be drawn to a second point above the horizontal, the i angle thus formed is the angle of elevation of the second point. If BC is a horizontal or level line the angle B is the angle of elevation of point A from B. 292. ANGLE OF DEPRESSION. If at any point in a horizontal line a line be drawn to a second point below the horizontal the angle thus formed is the angle of de- pression of the second point. L DAB is the angle of depression of B from A. 293. HORIZONTAL ANGLES. Angles in a horizontal or level plane are horizontal angles. 294. VERTICAL ANGLES. Angles of elevation and de- pression are vertical angles as distinguished from hor- izontal angles. 5. From the top of a tower the angle of depression of a vessel at sea is 15. If the tower is 87 ft. high hpw far away is the vessel? 124 PLANE GEOMETRY 1, Points B and C are on opposite banks of a river. Line BC along the bank is 150 ft. long, Z A is 43 , L C is 90. How Avide is the river ? 295. In any triangle the sides of two acute angles have the same ratio as he sides opposite. Given A ABC with sides a, b, c and acute angles B and C. A m _ sin B b To Prove - -. sin C c Sue. Express sin B and sin C. Complete the demonstration. NOTE: This is a very important proposition as it gives certain relations connecting the sides and angles of triangles other than right triangles. It is proved here only for the case in which the two angles concerned are acute. It will be proved in trigonometry without exception and is known as the law of sines. [2. In &ABC, A =15, C=76, and side c - 60 in. Find the other sides of the triangle and L B. sin A a , . . sin A SUG, 1. and .'. a c sin C c sin C 2. Since all angles of this triangle are acute this law may be likewise used for angles A and B. 3. Consult the table and complete the problem. 3. From a certain point the eleva- tion of the top of a church tower is 43. From a point 100 ft. nearer the base of the tower the angle of elevation is 55. Find height of the tower. Sue. Find sect BC from A ABC and then CD in A BCD. A- 4. Knowing the distance AC to be 8 mi., the angle A to be 40, and the angle C to be 82 j find the inaccessible distance AB across the lake. TRIGONOMETRIC RELATIONS 12s 296. TRIG. TABLE. The Sines, Cosines and Tangents of Acute Angles. Degrees Sin. Cos. Tan . Degrees Sin. Cos. Tan. 1 .0175 .9998 .0175 46 .7193 .6947 .0355 2 .0349 .9994 .0349 47 .7314 .6820 .0724 3 .0523 .9986 .0524 48 .7431 .6691 .1106 4 .0698 .9976 .0699 49 .7547 .6561 .1504 5 .0872 .9962 .0875 50 .7660 .6428 .1918 6 .1045 .9945 .1051 51 .7771 .6293 .2349 7 .1219 .9925 .1228 52 .7880 .6157 .2799 8 .1392 .9903 .1405 53 .7986 .6018 .3270 9 .1564 .9877 .1584 54 .8090 .5878 .3764 10 .1736 .9848 .1763 55 .8192 .5736 .4281 11 .1908 .9816 .1944 56 .8290 .5592 .4826 12 .2079 .9781 .2126 57 .8387 .5446 .5399 13 .2250 .9744 .2309 58 .8480 .5299 .6003 14 .2419 .9703 .2493 59 .8572 .5150 .6643 15 .2588 .9659 .2679 60 .8660 .5000 .7321 16 .2756 .9613 .2867 61 .8746 .4848 .8040 17 .2924 .9563 .3057 62 .8829 .4695 .8807 18 .3090 .9511 .3249 63 .8910 .4540 1.9626 19 .3256 .9455 .3443 64 .8988 .4384 2.0503 20 .3420 .9397 .3640 65 .9063 .4226 2.1445 21 .3584 .9336 .3839 66 .9135 .4067 2.2460 22 .3746 .9272 .4040 67 .9205 .3907 2.3559 23 .3907 .9205 .4245 68 .9272 .3746 2.4751 24 .4067 .9135 .4452 69 .9336 .3584 2.6051 25 .4226 .9063 .4663 70 .9397 .3420 2.7475 26 .4384 .8988 .4877 71 .9455 .3256 2.9042 27 .4540 .8910 .5095 72 .9511 .3090 3.0777 28 .4695 .8829 .5317 73 .9563 .2924 3.2709 29 .4848 .8746 .5543 74 .9613 .2756 3.4874 30 .5000 .8660 .5774 75 .9659 .2588 3.7321 31 .5150 .8572 .6009 76 .9703 .2419 4.0108 32 .5299 .8480 .6249 77 .9744 .2250 4.3315 33 .5446 .8387 .6494 78 .9781 .2079 4.7046 34 .5592 .8290 .6745 79 .9816 .1908 5.1446 35 .5736 .8192 .7002 80 .9848 .1736 5.6713 36 .5878 .8090 .7265 81 .9877 .1564 6.3138 37 .6018 .7986 .7536 82 .9903 .1392 7.1154 38 .6157 .7880 .7813 83 .9925 .1219 8.1443 39 .6293 .7771 .8098 84 .9945 .1045 9.5144 40 .6428 .7660 .8391 85 .9962 .0872 11.4301 41 .6561 .7547 .8693 86 .9976 .0698 14.3007 42 .6691 .7431 .9004 87 .9986 .0523 19.0811 43 .6820 .7314 .9325 88 .9994 .0349 28.6363 44 .6947 .7193 .9657 89 .9998 .0175 57.2900 45 .7071 .7071 1 .0000 126 PLANE GEOMETRY PROPOSITION XII. 297. THEOREM. In the same circle or in equal circles central angles are proportional to the arcs they intercept. Given OC = OC" with central angles C and C" in- tercepting the arcs AB and A'B' respectively. Z C arc AB To Prove Z C' arc A'B' Z C Proof. SUG. 1. To obtain the ratio the ZG" angles must be measured. Auth.? Suppose the unit angle e to be contained in Z C exactly m times and in Z C' exactly n times. 2. ... ^ = Why? Z C' n 3. To obtain the ratio the A'B' arcs must be measured. For a unit arc take the arcs intercepted by the unit angles. Auth ?. 4. How many of these unit arcs in ABt Why? In A'B' 6 ! Why? 5. What then is the ratio of AB to A'B"! / f1 4 "D 6. Compare the ratios - - and - L C' A'B' Therefore PROPORTIONAL LINES 127 NOTE. This demonstration does not cover the case in which the two angles are incommensurable. See note on 251. 298. DEGREE OP ARC. The arc intercepted by a cen- tral angle of one degree is a degree of arc. 299. COR. The number of degrees of angle in a central angle equals the number of degrees of arc in the intercepted arc. For: If central angle A intercepts arc a then 297 LA .__ arg a I 1 But these two ratios are respectively 1 of arc the number of degrees in the angle and the arc. This important theorem is usually stated thus: A central angle is measured by its intercepted arc. 297. If A and a are notations for the angle and arc the relation between them expressed in symbols is Z .4 = area. (50.) 1. Are all degrees of arc the same length! 2. How many degrees in a circle'? In a semicircle? Jn a quadrant ? PROPOSITION XIII. 300. THEOREM. An inscribed angle is meas- ured by one half its intercepted arc. Given O C with the inscribed Z ABD intercepting the arc AD. To Prove Z ABD =c \ arc AD. 128 PLANE GEOMETRY Proof. There are three cases. CASE I. One side of the angle, BD, is a diameter. SUG. 1. Connect A and C. Z m 2 Z B. Why? 2. .' LB = \ Zm. 3. ' Z n: I arc^LZ). Why? CASE II. The center of the circle lies between the sides of the angle. Sue. 1. Z m is measured by what? Why? 2. Z ?i is measured by what? Why? 3. L 'R is measured by what ? Why ? CASE III. The center of the circle is without the angle. Proof is left to the pupil. Therefore NOTE: This theorem may also be stated thus: The numbei of degrees in an inscribed angle is one-half the number in the intercepted arc and the same change in wording may be made throughout the proof. 301. COR. I. An angle inscribed in a semicircle is a right angle. 302. COR. II. A segment in which a right angle is inscribed is a semicircle. 303. COR. III. All angles inscribed in the same or equal segments are equal. 304. COR. IV. An angle inscribed in a segment greater than a semicircle is acute: in a segment less than a semicircle is obtuse. 1. Construct a right triangle by means of Cor. I. 2.1 In O C with diameter AB prove / m = L n. 3. If DE is a diameter of C which angle is the greater, w or n? n or of PROPORTIONAL LINES 129 PROPOSITION XIV. 305. PROBLEM. To construct a right triangle when the hypotenuse and an acute angle are given. A 'B A" Given sect AB as the hypotenuse and L A as an aciite angle of a right triangle. To Construct the triangle. SUG. Construct the given angle at point A of AB. At B erect a 1. Complete the problem. 1. To construct a perpendicular to a given line c from a point M on that line. See 301. Sue. With a point C not on the line as center and with CM as radius draw a circle cutting line c in a second point A. Draw AC meeting the circle again in B. Then MB is the required line. Why? 2. From a given point B not on a line c draw a perpendicular to c. See 301. Sue. Draw any oblique line through B meeting line c in point A. On AB as diameter, construct a circle meeting c in point M. The required line is MB. Why? 3. Draw two lines from A and B N meeting on the line MN so as to form a right angle. Is more than one construction possible? 4. Measure the height of your school building using two angles of elevation and n distance. See Ex. 4, P. 124. 5. With the four vertices of a square as centers and with radii equal to one-half a side of the square draw four circles. Show that one circle can be drawn which is externally tangent to the four circles. 130 PLANE GEOMETRY PROPOSITION XV. 306. THEOKEM. An angle formed by two in- tersecting chords is measured by one half the sum of the intercepted arcs. Given O O with chords CE and BD intersecting at A. To Prove Ln^\ (arc BC + arc DE). Proof. SUG. 1. Draw CD. L n = L C + Z D. Why? 2. "What is the measure of Z (7? of of ZD? 3. Complete the proof. Therefore 1. Show two methods of finding the center of an equilateral triangle. NOTE: The center of a polygon is that point which is equidis- tant from the vertices. Some polygons do not have centers. 2. On top of a hill '200 ft. high is a tower. From one point in the level plane at the foot of the hill the elevation of the top of the tower is 29. At a point 200 ft. nearer the foot of the kill the elevation is 40. How high is the tower? 3. Inscribe a triangle in a given circle similar to a given triangle. 4. Circumscribe a triangle about a given circle similar to a given triangle. 5. If two circles are tangent internally and through th point of tangency a line is drawn, the chords intercepted by the circu-s are proportional to the radii of the circles. PROPORTIONAL LINES 181 PROPOSITION XVI. 307. THEOREM. An angle formed by a tan- gent and a chord drawn from the point of contact is measured by one half the intercepted arc. Given the O with L m formed by the tangent AB and the chord AC. To Prove Z ra m J arc AEC. Proof. SUG. 1. Through C draw a line parallel to AB, as CD. 2. Compare A m and n ; arcs AEC and AFD. Auth. 3. Complete the proof. Therefore 1. Two circles are tangent internally. Two lines are drawn from the point of tangency through the extremities of a diameter of one circle. Prove that they intersect the other circle in the extremities of a diameter. 2. Prove Ex. 1 above if the circles are tangent externally. 3. AB and CD are two chords of a circle intersecting in the point 0. Prove A AOD and A COB mutually equiangular. Prove the same for A AOC and A BOD. 4. What is the locus of the vertex of the right angle of a right triangle with a given sect as hypotenuse? (302.) 5. Construct a triangle having a given base, a given altitude, and a right vertex angle. 6. Prove Prop. XV from the accom- panying figure in which EM \ BD. 132 PLANE GEOMETRY PROPOSITION XVII. 308. THEOREM. An angle formed by two secants, a secant and a tangent, or by two tan- gents is measured by one half the difference of the intercepted arcs. Given O with I. two secants AD and AE ; II. secant AD and tangent AB III. two tangents AB and AC. To Prove I. Z A m * (Arc DE etrcBC) ; II. LA^\ (Arc BD arc EC) III. Z A m | ( arc SMC arc BNC ) . Proof. CASE I. SUG. 1. Compare Z A with Z m and Z #. 2. What is the measure of Z m, Z #, Z A ? CASE II. Left to the pupil. CASE III. Left to the pupil. Therefore 1. All angles inscribed in the same segment are equal. 2. In the same or in equal circles an angle inscribed in the smaller of two segments is greater than an angle inscribed in the larger segment. 3. What is the locus of the vertex of a triangle having a given base and altitude? 4. A chord is met at one extremity by a tangent forming an angle of 75. How many degrees in the arc that is subtended by the chord? PROPORTIONAL LINES 133 PROPOSITION XVIII. 309. THEOREM. An angle formed by two lines either or both of which may be a secant or a tan- gent to a circle is measured by one half the sum of the intercepted arcs. Proof. SUG. 1. Is the theorem true if the lines in- tersect at the center ? Why ? 2. Is the theorem true if the intersec- tion be anywhere within the circle? Why? 3. If the intersection points ap- proach nearer and nearer to the circle what happens to one of the intercepted arcs? Sup- pose that when the point of intersection is on the circle this arc be considered as zero. Is the theorem true for this case? 4. A comparison of the figures shows that as moves from a position within the O to a position on the circle the points 0, A and C come together and as passes without the circle they again separate but with this dif- ference, that A and C, points in which the lines BA and DC meet the circle, in the third figure have their relative positions reversed, so that the arcs AC in the first and third figures are opposite in direction and if the idea of positive and negative lines be introduced the arc CA which enters the formula for the third case may be written as + AC. 269. 134 PLANE GEOMETRY 5. For each case then the measure of the angle at is J (arc DB + arc AC). The pupil should complete the demonstration for the case involving one or two tangent lines. 310. CONTINUITY. A theorem can sometimes be stated in such a general way as to cover two or more particular theorems. Especially is this true in geometry when a distinction is made as to the direction of lines represented algebraically by the use of positive and nega- tive quantities. Such a theorem is that of 309 which is proved by the principle of continuity. 1. In making a pattern for a certain casting it is necessary to construct a true semicircle. How may this be done witk a carpenter 's square ? 2. At a given point on a circle construct a tangent to the circle. Sue. If a line were drawn from the center C to the point A what relation would it bear to the tangent through A? Complete the construction. PROPOSITION XIX. 311. PROBLEM. From a given point without a circle to construct a tangent to the circle. Given O C with point A without. To Construct a tangent to O C from A. SUG. 1. Connect C and A. The problem is to determine the point of tangency. PROPORTIONAL LINES 135 2. What relation exists between the tangent and the radius at the point of tangency ? 3. What is the locus of a point with respect to AC satisfying this condition? 4. Complete the construction. 302. 5. How many such tangents are there? Why? 6. Make a second construction drawing only such portions of the auxiliary lines as are necessary. 312. FIND THE Locus : (1) Of centers of circles having a given radius and tangent to a given line. (2) Of centers of circles tangent to a given line at a given point. (3) Of centers of circles having a given radius and tangent to a given circle. (4) Of centers of circles tangent to two inter- secting lines. (5) Of centers of circles having a given radius and passing through a given point. (6) Of centers of circles passing through two fixed points. 1. Prove Prop. XVII by drawing from the point D in the accompanying figures a line DM 1 1 AC. 2. Two circles intersect in the points A and D. Lines AB and AC are diameters. Prove that B, D, and C lie in a straight line. 136 PLANE GEOMETRY 1. The sum of the distances from points in the base of an isosceles triangle to the legs is constant. 2. Given three non-parallel lines unlimited in length. Find points in one of them equally distant from the two others. How many such points are there? 3. Two circles 'intersect in the points A and D. Sects AB and AC are chords of the two circles. Points B, I), C are in a straight line. Prove that the chords are diameters. 4. One side of an equilateral inscribed hexagon is equal to the radius of the circle. 5. Inscribe an equilateral triangle in a circle and prove that the radius perpendicular to a side is bisected by the side. Sue. OABD is a /~7. Why? 6. Two chords drawn from a point on a circle are inversely proportional to the seg- ments of the chords included between the tan- gent at their common point and a line parallel to this tangent. To PROVE * = 4*.. AE AC See note, 285 (1). 7. The locus of the middle points of all chords which pass through a given point is a circle the diameter of which is the line joining the given point and the center of the given circle.. Sue. Prove that the circle described on DC is the required locus. 8. What has been done to the proportion = to produce n s s o+s s 9. Construct a circle having a given radius through a given point and tangent to a given line. SUG. Find two loci of the centers. PROPORTIONAL LINES 137 PKOPOSITION XX. 313. THEOREM. // two chords intersect the ratio of either segment of the one to either seg- ment of the other is equal to the ratio of the remaining segment of the second to the remain- ing segment of the first. Given a circle with chords AB and CD intersecting at E. AE CE AE DE To Prove = or = DE BE CE BE Proof. SUG. 1. As no A are given in the theorem, two A must be constructed one of which contains the required antecedents and the other the re- quired consequence. Note 285 (1). 2. Prove the constructed triangles similar. 3. Establish the required proportions. Therefore Make a second construction and proof for this proposi- tion. 314. COR. The product of the segments of one of two intersecting chords of a circle equals the product of the segments of the other. To Prove AE*EB = CE* ED. 1. The line joining the center of a circle with an outside point bisects the angle made at that point by the two tangents to the circle. 138 PLANE GEOMETRY PBOPOSITION XXI. 315. THEOREM. // two secants intersect with- out a circle, the ratio of the first to the second is equal to the ratio of the external segment of the second to the external segment of the first. Given a circle with secants AB and AC intersecting the circle in the points D and E respectively. To Prove =^ A C AD Proof. The desired conclusions will at once follow if two triangles can be constructed, one of them contain- ing the required antecedents and the other the required consequents. Try such a construction and complete the demonstration. Note 285 (1). Therefore 316. COR. // two secants meet without a circle the product of one secant and its external segment equals the product of the other and its external segment. 1. In a given circle two radii are drawn to the center of two chords, and the angles made by them with the sect joining their feet upon the chords are equal; prove the choids are equal. 2. What line does the center of a ladder discribe as its foot is drawn directly away from the building against which it leans, the ground beijig level. 3. What is the locus of the center of a given sect whose end points continually touch the respective side of a right angle'? Suo. Ex, 2. PROPORTIONAL LINES 139 317. THEOREM. In a series of equal ratios the ratio of the sum of the antecedents to the sum of the conse- quents equals any of the given ratios. -. abed Given -=-- = -=-. etc. a b c d a +6 +c +cZ + . a To Prove - = a'+6'+c'+d'+... a' Proof. SUG. 1. Let n represent the common value of the given ratios. Then a = na', b = nb', etc. 2. .-. a -I- b + c + d + . . . = n (a f + V + c'+d' + ...) a + b + c + d + ... a 3. .-. = M = -- a'+ b'+ c'+ d'+ ... a' 4. Give the authority for each step and verify by a particular set of numbers. Therefore 1. Give a summary of the tests for similarity of triangles. 2. Through a given point draw a line so that the sect in- tercepted between two given intersecting lines is bisected at the given point. SUG. Through the given point, o, draw a line parallel to one of the given lines. 3. Through a point included be- tween the sides of an angle draw a line so that the sect intercepted be- tween the sides shall be divided in the ratio of 1 to 4. SUG. Draw MN\\AB. Lay off on side AC a sect MP equal to 4AM. Draw PO and extend to AB. 4 The sum of the three perpendiculars in the preceding ex- ample equals the altitude of the triangle. 5. The three altitudes upon the three sides of an equilateral triangle are equal. 6. The sum of the perpendiculars from any point in an equilateral triangle to the three sides is constant. 140 PLANE GEOMETRY PROPOSITION XXII. 318. THEOREM. The ratio of the perimeters of two similar polygons equals the ratio of simil- itude of the polygons. Given two similar polygons P and P' with sides a, b, c, . . . and a', &', c', . . . respectively, and perimeters p and p' respectively. To Prove =*- p' a'- Proof. SUG. 1. Write out in detail the relation in- volving the sides of P and P'. 2. Express the ratio in terms of P' the sides. 3. Complete the proof. Therefore 319. MEAN PROPORTION. A proportion in which the means are the same is a mean proportion. 320. MEAN PROPORTIONAL. The second or third term in a mean proportion is a mean proportional. 321. THIRD PROPORTIONAL. The fourth term of a mean proportion is the third proportional. a _b ^ ~ is a mean proportion. The mean proportional is b and the third proportional is c. 1. Inscribe in a circle an isosceles triangle with a diameter of the circle as its base. How many degrees in the vertex angle? 2. If a right triangle is inscribed in a circle, the hypothenuse is always a diameter. 3. A line is drawn from the vertex of a triangle to the base. What is the locus of a point that divides it always in the same ratio? PROPORTIONAL LINES 141 PEOPOSITION XXIII. 322. THEOREM. A mean proportional of two num- bers equals the square root of their product. -. a b Given - = b c To Prove b = V~ac. Proof. ac = b' 2 . Why? Whence 6 =? 4 x 1. I* x = ~9 find *' ii. li 3 x = find x. x / 3. If 7- = 7 find a;. c> iC PROPOSITION XXIV. 323. THEOREM. // /ie altitude on the hypot- enuse of a right triangle be drawn.- I. the triangle is divided into two triangles, each similar to the given triangle and similar to each other; II. the altitude is a mean proportional to the segments of the hypotenuse; III. each side of the given triangle is a mean proportional between the whole hypotenuse and the adjacent segment. A MB Given A ABC with altitude MC drawn to the hypote- nuse AB. To Prove I. A AMC ^ A A CB, A CMB <-" A ACB, A AMC "" A CMB. 142 PLANE GEOMETRY II. M C a mean proportional to AM and AM MC MB, i. e. = MC MB III. AC a mean proportional to AB and AB AC AM, i. e. - = or BC a mean proportional to AB AC AM AB BC and MB, i. e. = BC MB Proof. I. SUG. L A is common to & ^LC and ACS. ' A AMC"-* A AC. Why ? Complete the proof of I. II. SUG. In &AMC and CMJ5 sides AM and . (7 are homologous, also sides MC and MB. Why? Complete the proof. Apply carefully 285 ( 1 ) note. The difficulty in this case lies in the fact that MC is in two different triangles. III. SUG. Use A AMC and ACS, noting the equal angles. Select the homologous sides for each ratio in accordance with the suggestion of 285 (1). Therefore 324. COR. In a semicircle a perpendicular from the arc upon the diameter is a mean proportional to the seg- ments of the diameter. NOTE: In reading similar polygons care should be observed to follow a reading which gives the same order to homologous parts. 1. A perpendicular dropped from a circle upon a diameter is a mean proportional between the segments of the diameter. 2. Use the preceding exercise to construct a mean propor- tional to the sects a and b. EXERCISES 143 3. By means of this proposition. Til, construct a mean propor- tional to two given sects. Also a third proportional. 4. If the altitude upon the hypotenuse divides the hypotenuse into two segments of 4 and 6 in. respectively, what is the length of the altitude? What is the length of each side? Express results accurate to nearest integer. One segment is 8 in. and the adjacent side is 12 in. Find the hypotenuse and the two other sides. 5. Of all parallelograms having equal bases and altitudes the rectangle has the least perimeter. 6. The diagonals of an isosceles trapezoid are equal. 7. If two parallel straight lines are cut by a transversal, the bisectors of the interior angles on the same side of the transversal are perpendicular to each other. 8. If the median drawn from the vertex angle of a triangle to the base equals half the base, the vertex angle is a right angle. Of which exercise is this the converse? 9. If a side of a triangle is parallel to the bisector of an exterior angle of the triangle, the triangle is isosceles. 10. In physics one learns that the angle of incidence equals the angle of reflection. In the figure these angles are i and r respectively. Prove that an object viewed in the mirror M seems as far behind the mirror as it is in front of it. observer object 11. The bisectors of the angles of a parallelogram form a rectangle. 12. a, b, c are the angles of an inscribed triangle. L a is four times b and b is one seventh of c. How many degrees in the arcs of the circle subtended by the respective sides of the triangle? 13. In the figure of Prop. XVI draw the diameter AM and prove the theorem. 144 PLANE GEOMETRY PROPOSITION XXV. 325. PROBLEM. Construct a mean propor- tional to two given sects. Given sects a and b. a T To Construct sect x so that - = Construction. SUG. 1. Make a drawing of the figure for Case II, Prop. XXIY to see which lines must be taken for a, b, x. Then construct the figure from the sects a and b. 2. Or in a similar manner make use of figure and conclusion of ex. 1 p. 142. 1. Construct a mean proportional to two given sects a and b by use of Case III of Prop. XXIV. 2.. The bisectors of all angles inscribed in the same segment pass through a common point. Where is that point? 3. Draw any two equal non-intersecting chords of a circle and connect their adjacent extremities. Prove that the connecting chords are parallel. 4. Draw any two parallel chords and connect their extremi- ties. Prove that the connecting chords are equal. 5. ABC is a triangle inscribed in a circle with center 0. Lino OD is perpendicular to BC. Prove L DOC, or its supplement, equal to ZA. PROPORTIONAL LINES 145 PBOPOSIT10N XXVI 326. THEOREM. // from a common point a tangent and a secant be drawn to a circle, the tangent is a mean proportional between the secant and its external segment. Given a tangent AB and a secant AE of circle C. _ AD AB To Prove - = --- AB AE Proof. SUG. 1. Indicate a triangle having AE and AB as sides. Do the same for AB and AD. These 4 are similar. Why? 2. Complete the demonstration. Therefore 327. EXTREME AND MEAN RATIO. A sect is divided into extreme and mean ratio when the greater segment is a mean proportional to the whole sect and the smaller segment. AB is divided internally at M in extreme and mean ratio if AB AM ~A~M = llfB an ^ externally at M' in extreme and mean ratio if AM' ~ M'B B A w PEOPOSITION XXVII. 328. PROBLEM. To divide a sect internally and externally into extreme and mean ratio. Given sect AB. To divide AB at M and at M 1 so that = and 146 PLANE GEOMETRY AB AM' AM' M'B '~if n respectively. B M A ~~M' SUG. 1. Erect a perpendicular at B (or at A) equal to | AB. With the free extremity of this perpendicular as a center and OB as a radius describe a circle. Connect A and and extend the join line to meet the circle again at C. Take 2. ^ = ^ = L:. Why? AB AD AM 3. Form a new proportion from thk by division and complete the proof. 4. Therefore M is the required point of internal division. II. SUG. 1. Extend BA to M' so that AM'=AC. 2. ^ = ^. Why? AC AB 3. Form a new proportion from this by composition and complete the proof. 4. Therefore M' is the required point of external division. 329. It must be remembered in the series of theorems follow- ing that in the operations of multiplication, division, squaring, etc., the symbols represent numbers, viz: The measures of the lines of the same name. Sect a X sect & means the product of the numerical measures of sect a and sect &. For brevity the operation of measuring will be assumed to have been done and the name or notation of sect will also represent the measure of it. NUMERICAL RELATIONS 147 The operations performed in the demonstrations are then those of algebra. 252. 1. If the hypotenuse is divided by the altitude from the right angle, the ratio of the squares of the sides equals the ratio of the adjacent segments of the hypotenuse. a z m To PROVE p = (Use fig. of Prop. XXIV.) Sue. a 2 = me. Why? b- --/?" Complete the proof. PROPOSITION XXVIII. 330. THEOREM. The square of the hypote- nuse of a right triangle equals the sum of the squares of the two other sides. Given a right A with sides a and b and hypotenuse c. To Prove a 2 + 6 2 = c 2 . Proof. Sue. 1. a 2 = mc. Why? & 2 =? 2. a 2 + b 2 = ? Notice that m and n are here coefficients of c. Therefore Another proof is found in 376. 331. COR. I. The square of either side of a right triangle equals the difference of the squares of the hy- potenuse and the other side. 332. COR. II. The ratio of a diagonal TO the side of v~2 a square is : = V2 2. If the side of a square is 4 in., what is the diagonal! Find the result correct to two decimal places. 3. One side of a rectangle is 4 in. and a second side is 9 in. Find the diagonal correct to two decimal places. 148 PLANE GEOMETRY 4. A carpenter squares the frame of a building by measur- ing 6 ft. from the corner on one side, 8 ft. from the corner on the other, and then draws the frame until the distance across from the ends of the sects is 10 ft. Verify the correctness of his plan. 5. The distance from A to B is desired. A distance 70 rods is measured from B to C, so that the angle BAG is 90. If AC is found to be 60 rods, how long is AB ? 6. The diagonal of a rectangle is 12 in. Find the sum of the squares of the four sides. 7. Two sides of a rectangle are 6 and 8. Find the diagonal. Do the same for sides 5 and 12; 6 and 11. 8. Construct a right triangle and the altitude upon the hy- potenuse. Measure the segments of the hypotenuse and compute the altitude. Check your work by measuring the altitude. 0. Find a mean proportional to 3 and 5; to 4 and 7; to G and 11. Use algebra and also geometric construction. 10. If # = V 3 X 4 find x. Sug. x is a mean proportional to 3 and 4. Why? Construct x geometrically and verify by measurement. 11. Find a line the length of which is V~12~. SUG. VTU = V3 x 4. Make two other constructions. Use 324. 12. Find v/~8~ by the measurement of a line. Also V~57 VT^ Vl5. Sue. Use 324 and verify results by extracting the roots. 13. A boy in traveling from his home A to a town B on his wheel traveled 16 mi. in a straight line and then 24 mi. in a line at right angles to the first road. How many miles would he have saved by traveling along the railroad which ran in a straight line from A to U? 14. The sum of the squares of the two diagonals of a rectangle equals the sum of the squares on the four sides. 15. Find the length of the side of a square correct to two decimal places if the diagonal is 1. If the diagonal is 18; 2; 32; 26; 4; 12. 16. Find the diagonal of a square the side of which is 8. Do the same if the side is 25; 3.15; 4.18. NUMERICAL RELATIONS 149 PROPOSITION XXIX. 333. THEOREM. The square of the sum of two sects equals the sum of the squares of the sects plus twice the product of the sects. Sue. If a and b are the measures of the two given sects, 329, it is necessary and sufficient to prove the formula (a + b) 2 = a 2 + b 2 + 2ab. The pupil should do this by performing the indicated multi- plication. Another proof is found in P. 186, PROPOSITION XXX. 334. THEOREM. The square of the difference of two sects equals the sum of the squares of the sects minus twice the product of the sects. Given a and b as the measures of the two sects. To Prove (a b) 2 = a 2 + b- 2 ab. Proof left to the pupil. 335. PROJECTION OF A POINT. The A projection of a point upon a line is the foot of the perpendicular drawn from the point to the line. 3 If AM is perpendicular to EC, then M is the projec- tion of A upon BC. 336. PROJECTION OF A SECT. The projection of a straight sect upon a straight line is the sect between the projections of the extremities of the given sect upon the same line. 1. The product of the sum of two sects by their difference equals the difference of the squares of the sects. To PROVE (a -I- 6) (a 6) = a? 2. M 150 PLANE GEOMETRY PROPOSITION XXXI. 337. THEOREM. The square of the side oppo- site an acute angle of a triangle equals the sum of the squares of the two other sides minus^the product of one of those sides and the projection of the other upon it. Given a A with the sides a, b, c, side a being opposite an acute angle and m being the projection of b upon c. To Prove a 2 = b 2 + c 2 2mc. Proof. SUG. 1. Drop a perpendicular, p, from ver- tex C to side c. 2. a 2 =p 2 +w 2 = p 2 + (c-m)* = &2 - m 2 + ( c - m) 2 = b 2 - m 2 + 3 2 + m 2 2wir == 62 + c 2 2wc. Therefore 1. If a quadrilateral is circumscribed about a circle, the sum of one pair of opposite sides is equal to the sum of the other pair. 2. If two circles are tangent and two secants are drawn through the point of contact the two chords joining the points in which the secants meet the respective circles are parallel. PROVE AD | BC. Sue. Draw the common tangent MN. Compare & AOM and CON ; A AOM and ADO; A. OBC. NUMERICAL RELATIONS 151 PROPOSITION XXXII. 338. THEOREM. The square of the side oppo- site an obtuse angle of a triangle equals the sum of the squares of the two other sides plus twice the product of one of those sides and the projec- tion of the other upon it. Given a A with sides a, b, c, side a being opposite an obtuse angle, and m being the projection of b upon c. To Prove a 2 = b 2 + c 2 + 2mc. Proof. Suo. 1. The projection m will in this case lie on the extension of c. 2. Complete the proof in a manner similar to the method of 337. 339. The theorems of 337 and 338 may be in- cluded in one statement which may be proved by the principle of continuity ( 310) as soon as a distinction is made as to the direction of the line representing the projected side. The theorem may be stated in general as follows : The square of any side of a triangle equals the sum of the squares of the two other sides minus twice the product of one of those sides and the projection of the other upon it. c c - 152 PLANE GEOMETRY Given a A with sides a, b,c, LA being acute or obtuse, or right, and m, the projection of h upon c, be- ing considered as positive when it extends in the direc- tion AB and negative when it extends in the opposite direction. To Prove a 2 ~= 6 2 + c 2 2wc. Discussion. As angle A increases from an acute an- gle to an obtuse angle, passing through the value 90, point M, the projection of vertex C upon side c moves towards the vertex A, passes through this point when Z A 90, and passes out on the extension of side c through A when Z A becomes obtuse. For Z A acute, m is positive ; for Z A = 90, m is 0; and for Z A obtuse, m is negative. The formula a 2 = 6 2 + c 2 2wc for these three values of m reduces to the formulas previously proved for these three cases in 337, 330, and 338 respectively. This theorem will be met again in trigonometry under the name of The Law of Cosines. 1. Fasten a rubber band to the two points of a pair of di- viders. Let the rubber band represent side a in the triangles of 339, and the arms cf the dividers the sides b and c. The angle made by the arms then represents angle A. As the dividers are slowly opened, note the change of LA from acute values to 90 and then to obtuse values and at the same time observe the shortening of the projection of & on c from a positive sect through zero on to increasingly large negative sects. 2. The sum of the squares of the diagonals of a parallelogram equals the sum of the squares of the four sides. 3. The sum of the squares of the diagonals of a quadrilateral equals the sum of the squares of the four sides plus four times the square of the sect joining the mid-points of the two diagonals. 4. A chord of a circle, AB, is 5 in. long. If it be produced to a point C so that the external segment BC is 10 in., how long is the tangent from C? NUMERICAL RELATIONS 153 PROPOSITION XXXIII. 340. PROBLEM. To find any altitude of a tri- angle in terms of its sides. Given A with sides a, b, c, and h a the altitude upon a. To find h a in terms of a, b, c. SUG. 1. h a * = & 2 ZJC 2 and c 2 a 2 +6 2 2aX/>C. 2. . . DC = 2a 3 . , (c+a-b) (a+b-c) (a+b+c) 4. Let 2* = 2(s a) c a + &, 2(s b) c+a b, 2(sc) = a + b c, whence, 2 ^ 2 -^- 5. Similarly /< , = V (sa)(sb)(sc)s b 2 AV = - V (sa)(sb)(sc)s c 1. By formula, find the three altitudes of triangle the sides of which are 12, 14, 18. Construct the triangle and measure the 154 PLANE GEOMETRY altitudes as a check on the computations. Do the same for the triangle with sides 7, 9, and 12; also 13, 15, and 20. PROPOSITION XXXIV. 341. THEOREM. In any triangle the product of any two cides equals the product of the alti- tude upon the third side and the diameter of the circumscribed circle. I Given A ABC with sides a, b, c, altitude h a upon side a and AM or d the diameter of the circumscribed circle. To Prove bc = dh a . Proof. SUG. 1. Connect M and B. A CAD and MAB are similar. Why? 2. Deduce a proportion from which the required equality may be obtained. Therefore PROPOSITION XXXV. 342. PROBLEM. To find the length of the radius of the circumscribed circle in terms of the sides of the triangle. Given A ABC with sides a, b, c, altitude h a upon side a, and AM or d the diameter of the circumscribed circle. To find d in terms of a, ft, c. (See fig. of Prop. XXXIV,) Sue. 1. h a xd = bc. Why? NUMERICAL RELATIONS 155 3. Substitute for k a its value in terms of a, b, c. 4. Simplify this result, obtaining abc 4Vs(s a) (sb) (sc) 1. Find the radius of the circle circumscribed about a tri- angle with sides 8, 9, and 12. Construct the figure and check the computation by measurement. PROPOSITION XXXVI. 343. THEOREM. In any triangle the square of the bisector of any angle equals the product of the sides including the angle minus the product of the segments of the third side made by the bisec- tor. c Given A ABC with BD bisecting L B, and meeting AC at D. To Prove ~BD* = AB x BC AD X DC. Proof. SUG. 1. Circumscribe a circle and extend BD to meet the circle at M. Draw MC (or MA). 2. &ABD^&MBC. Why? 3. = Why? MB BC 4. . BD 5. .'. BD2 = AB x BC AD X DC. 6. Verify each step. Therefore 156 PLANE GEOMETRY PKOPOSITION XXXVII 344. PROBLEM. To find the length of the bi- sector of an angle of a triangle in terms of the sides of the triangle. D a Given A ABC with sides a, b, c, AD or d n being the bisector of Z A, meeting EC at point D. To find an expression for d a in terms of a, &, c. SUG. 1. By 343 d a * = bc- BDxDC. Also ##__c Z)6'~6 Why? c & c+& c+& r>/? n ]\ 3. b+c b+c ac ab a*bc = ?).^ x -- = be b+c b+c (b+c)* c _ be \_(b+c)* a 2 ]_ _bc(b+ca) (a+b+c) (b+c) 2 (b+c) 2 _ d a = - - Vbcsfba), in which 2.s = a+6+c, as in 340. 6. Verify each step and write corre- sponding expressions fcr d fi and r/ . NUMERICAL RELATIONS 157 1. If the sides of a triangle are 7, 9, and 12; find the length of the bisector of each angle. Construct the triangle and check the computations by measurements. 2. In triangle ABC side a = 24, & = 17, and c = 20; find the bisector of L C. 3. Find the radius of the circle circumscribed about a tri- angle with sides of 12, 18, and 20. Check the results by con- struction. PROPOSITION XXXVIII. 345. THEOREM. In any triangle if a median be drawn to the base the sum of the squares of the two other sides is equal to twice the square of half the base plus twice the square of the median. D Given a A with sides a, b, c, with D the projection of C and d the projection of m c on c, and m c the median on side c. To Prove a 2 + 6 2 = 2m 2 + 2 'U Proof. SUG. 1. If a and b are unequal, one angle between m c and c is acute and the other is obtuse. Suppose then that a is opposite an obtuse angle. 2. Express a 2 and b 2 in terms of the remaining sides of the two triangles made by m c . Add the resulting expressions. 3. Prove the proposition when sides a and b are equal. Therefore 158 PLANE GEOMETRY 1. Prove 343 for the bisector of an exterior angle. Sue. Letter the figure as in the original proposition. Notice that AC is divided externally at D and AC = AD CD. 2. Discuss the above problem when the angle bisected is exterior to the vertical angle of an isosceles triangle. In this case BD is parallel to AC. Why? No conclusions can be drawn as in the other cases. BD, AD, CD, all become infinitely large. PROPOSITION XXXIX. 346. PEOBLEM. To find the length of any me- dian in terms of the sides. Given A ABC with sides a, b, c and m a the median on side a. To express m a in terms of a, b, c. From2w? = b*+c*-2l - \ and Write the corresponding values of r in b and m c . *This proposition may be omitted. PROPOSITION XL. 347. THEOREM. In any triangle, if the median be drawn to the base, the difference of the squares of the two sides equals twice the product of the base and the projection of the median on the base Given the conditions of Prop. XXXVIII. To Prove a 2 b 2 = 2cd. Proof. Subtract b 2 from a 2 . Therefore EXERCISES 159 1. In &ABC side a = 8, & = 12, and c = 14. Find the length of the median on c. Construct the triangle and median and cheek the computations by measurement. 2. In the above example find m^ and m and check computa- tions. Given the sides a = 10, & 6, and c=8. Find the median on a and check the computation. 3. Draw a triangle and a median with a straight edge. Meas- ure the sides and the median. Check the results by computing the length of the median, and also by using other theorems in- volving the median and sides. 4. If two circles are tangent and a sect be drawn through the point of tangency terminated by the circles. I. Tangents to the circles at the extremities of the sect are parallel. SUG. Draw the common tangent at the point of tangency. Consider the two cases of internal and external tangency. II. The diameters of the two circles through the extremities of the sect are parallel. Sue. Use same construction as above, with two cases. 5. A ABC is equilateral and AB and AC are tangents to the circle. How many degrees in arc BC1 In arc BMC1 M( 6. If AB is 18 in., what is the diam- eter of the circle? 7. Show that OB is a mean proportional to OE and OA. OE _ OB To PROVE Q2 - 02- 318. EXTERNAL TANGENT. A line is an external tan- gent to two circles if it is tangent to each but does not cut the line of centers. INTERNAL TANGENT. A line is an internal tangent to two circles if it is tangent to each and cuts the line of centers. AB is an external tangent and CD an internal tan- gent. 160 PLANE GEOMETRY PROPOSITION XLI. 349. PKOBLEM. To construct a common ex- ternal tangent to two circles. Given two C and C'. To Construct AB a common external tangent to C r.nd C'. SUG. 1. Construct a circle concentric with the larger of the given circles with a radius equal to the difference between the given radii. Let this radius be CM. Why ? 2. Draw a tangent to this auxiliary circle from C' '. How may this be done? 3. Draw perpendiculars to this tan- gent line from C and C' meeting the circles in the points A and B respectively. Why ? 4. AB is the required external tan- gent. Explain why. 5. Is AB the only external tangent? In what step of the construction might a second one be introduced? 1. The common tangent to the two ex- ternally tangent circles at the common point meets an external tangent in a point equally distant from the three points of contact. To prove that OA = OB = OC. 2. If two circles are tangent externally the sects which join the point of contact to the two contact points of an external 1;m- gent form a right angle. To prove L ACB =90. PROBLEMS OP CONSTRUCTION 161 1. Two parallel tangents to a circle intercept a sect on a third tangent. Prove that this sect subtends an angle of 90 at the center. 2. If two circles are externally tangent the sect which is the common external tan- gent is a mean proportional to the diameters. 3. Construct a mean proportional to sects a and b by means of Ex. 2. PROPOSITION XLII. 350. PROBLEM. To construct an internal tan- gent to two circles. Given C and C". To Construct AB, an internal tangent to C and C' . SUG. 1. Draw a circle concentric with the smaller of the given circles and with a radius equal to the sum of the radii of the given circles. Why? 2. The completion of the proof is left to the pupil. 351. In the previous pages, certain principles of con- struction were stated and certain constructions made. In most of the problems thus far the pupil has been given more or less aid. In order that the pupil may become more independent in the solution of problems certain methods of analysis will now be studied in con- nection with the solution of a few problems. 162 PLANE GEOMETRY Constructions are based upon previously demon- strated theorems, and the difficult thing in the solution of any problem is the recognition of the previous propo- sitions which may be applied in effecting the desired construction. In the more difficult problems this can most easily be done by first making a sketch of the com- pleted construction. This construction should next be analyzed in order to discover the theorems involved; then by reversing the order of the analysis, the con- struction can usually be made. Each problem usually gives certain conditions for the determination of one or more points, the geometric representation of which conditions are loci, their in- tersections being the required points. The number of solutions, when a problem calls for the location of a point, is the same as the number of such intersections. In the case of no possible intersection, there is no solu- tion. 352. PROB. I. To construct a triangle, given one side, an adjacent angle, and the altitude upon that side. Given side AB, L A, and alti- tude MC. As AB is given it can be drawn and at one extremity the given an- gle A can be constructed. The vertex C must now be determined. One locus of C is the unlimited side of L A, i.e. line AC. The other locus of C must be found from the other condition, viz : the given altitude MC. It is evident that the locus of the vertex of a triangle with a given altitude is a line parallel to the base and at a distance from the base equal to the given altitude. The construction can now bo easily made. PROBLEMS OF CONSTRUCTION 163 Discussion: On a given side of line AB there is but one solution for two straight lines can have but one in- tersection. Also there is always a solution for if a line cuts one of two parallels it will cut the other. PROB. II. Upon a given sect to construct a segment of a circle which shall contain a given angle. Given Z n and sect BC. The problem is easily solved if the center of the required circle is found. c- One locus of the center is the perpendicular bisector of sect BC. Why ? it is ob- served from the figure that if the segment B A C con- tains Z A = L n that the arc BC is not only twice the measure of LA but also of the angle at B between CB and a tangent to the circle. In other words if an angle equal to Ln be constructed at B, the second side will be tangent to the required circle. The perpendicular to this tangent at B is also a locus of the center. The cen- ter is thus determined. Does this problem have a solu- tion for every sect BC and Z n ? PROS. III. To construct a circle with a given radius that shall pass through a given point and be tangent to a given circle. Given a circle C, a fixed point P, and a sect r. To Construct a circle with radius r, passing through P, and tangent to C. Draw a figure representing the completed construction. The center C' of the required circle is needed. What is the 164 PLANE GEOMETRY distance from C' to Cf What then is a locus of G'? Since the distance PC' is r, what is a second locus of C' ? Discussion: In. general how many intersections of these two loci and hence how many solutions ? Is it pos- sible for one solution only to exist? In this case how would the distance PC compare with the radii of the two circles? What relation of distances would make the in- tersection of the two loci an impossibility? Illustrate these special cases by figures. PROB. IV. To construct a triangle having given the base, the altitude, and the radius of the circumscribed circle. ^ To Construct a triangle ABC with b the sects a, ~b, and c as base, altitude, _ and radius of the circumscribed circle respectively. The problem is solved as soon as one finds the third vertex, A, the extremities B and C of sect a being two of them. What is the locus of vertex A as determined from the given altitude? The center of the circumscribed circle is at distance r from B and C. Locate its center 0. The circumscribed circle is the second locus of vertex A. Discussion. The first condition determines as a lo- cus for A two straight lines. Why two? The Second condition determines two cir- cles. Why? If the altitude is less than MD, each of the straight line loci meets each circle locus twice. How many solutions? If the al- titude is greater than MD EXERCISES 165 but less than MP, each straight line locus meets one circle locus twice. How many solutions? If the al- titude equals MD and is less than MP each straight line locus is tangent to one circle locus and meets the other in two points. How many solutions? If the altitude equals MP each straight line locus is tangent to one cir- cle and does not meet the other. How many solutions? If the radius equals one half the base, then MD = MP, the problem is impossible or there would be two or four solutions, and any of the above cases in which MD and MP are assumed unequal are impossible. Discuss this in detail. If the radius is less than half the base, there is no solution. Why ? 1. The locus of the vertex of a given angle subtending a given sect is the arc of the segment which contains the angle. SUG. 1. Construct the segment containing the given angle. 2. All angles inscribed in this segment equal the given angle. Why? 3. Any angle not inscribed in this segment and subtending the given sect is either greater or less than the given angle. Why! 2. Divide a sect 8 in. long internally into extreme and mean ratio. Measure the segments and arithmetically check the con- struction. 3. Divide 8 into two parts such that one part is a mean proportional between 8 and the other part. 4. Given a chord in a circle. From any point in the arc draw a second chord which shall be bisected by the first. Two solutions. 166 PLANE GEOMETRY 5. Construct a triangle, given the base, the altitnde, and the vertex angle. 0. Construct a triangle given two angles and a side opposite one. 7. Construct a triangle, given two angles and the altitude upon the included side. 8. Construct a triangle, given the base, the median to the base, and an angle adjacent to the base. 9. Construct a triangle, given the base, the median to the base, and the vertex angle. 10. Construct a circle of given radius and tangent to two given circles. What conditions are necessary for a solution? How many solutions? 11. Construct a circle having a given radius, tangent to a given line, and passing through a given point. How many solu- tions and what conditions govern each case? 12. Construct a mean proportional to two sects, using Ex. 1 p. 142. 13. Construct a third proportional to two sects. GIVEN sects in and n. m _ n To construct sect x such that ~ It Ju SUG. 1. On an un- limited line lay off a m sect m + n. Through m one extremity of this sect draw an oblique line and from the intersection lay off on the line sect n. Join the ex- tremities of the two sects of lengths m and n and through the free extremity of sect m + n draw a line parallel to this join line. The sect intercepted on the oblique line by the parallels is x. 2. Verify the construction. ft 14. Construct a fourth proportional to f sects a, I), c, using the method of Ex. 13. EXERCISES 167 15. Construct a fourth proportional to three sects using the problem for constructing a triangle similar to a given triangle. 16. Construct a fourth proportional to three sects using Ex. 1, p. 120. NOTE: For such a problem any theorem in which a propor- tion of four different terms has been established may be used. 17. If a chord is 12 in. long and the perpendicular to that chord bisecting the subtended arc is 3 in., find the radius. 18. Draw accurately the figure of the preceding exercise, con- struct the circle and check the computations by measurements. 19. A bridge with a circular stone arch has a 39 ft. span and an 8 ft. rise. What is the diameter of the circle? 20. The altitude of an equilateral triangle is 12 in. Find the length of a side. 21. Prove that the perpendiculars from the center of a circle to the sides of an inscribed equilateral polygon are equal. This perpendicular is the apothem of the polygon. 22. Circumscribe a circle about an equilateral triangle and compare the apothem with the radius and the altitude. 23. Given an equilateral triangle of side 12 in. Find the apothem and altitude. Do the same for a triangle of side a. 24. Connect the vertices of an inscribed equilateral triangle with the mid-points of the arcs between them. Prove that the resulting hexagon is equilateral and equiangular. 25. Prove that the side of an inscribed equilateral hexagon equals the radius. 26. The sides of a triangle are 8, 11, 13. Find the altitudes, medians, bisectors of the angles, and the radius of the circum- scribed circle. 27. In a circle of 10 in. radius there are two parallel chords, one being 6 in. from the center and the other 8 in. Find their lengths. 28. Find a fourth proportional to 2, 7, and 8 by arithmetic and geometric methods. Check the results by comparison. 29. Find a fourth proportional to 5, 7, and 9 by arithmetic method and by geometric construction. Check the results by comparison. 30. Find by arithmetic method and by geometric construction a sect equal to V~9; VT; VTO. Compare the results. 1f>8 PLANE GEOMETRY 31. If two circles intersect, tangents drawn from any point in their common chord extended, are equal. 32. CD is tangent to circle 0, AB is a secant cutting the circle at points A an B and perpendicular to CD. At what point X in CD is L AXB the greatest ? E 33. If A and B are goal posts on a fcot ball field and E the place of a touch down, how far back afield should the attempt be made to kick goal to insure the best oppor- tunity? 34. The goal posts are 18% ft. apart, and the touch down is made 49 feet from the nearest post. How many feet back is the best point from which to kick? 35. To find AB, the distance across the river. L A is found to be 107, ZG' to be 53, and line AC to be 24.7 rods. CHAPTER IV. AREA OF POLYGONS 353. AREA. Area is a species of quantity ( 245 ) and is obtained by measuring surface. 354. MEASUREMENT OF SURFACE. Measurement of surface is the process of determining the number of times a unit of surface is contained in the given surface. 355. UNIT OF SURFACE OR AREA. A square having a given linear unit for one side is usually taken as the unit of surface or area, e. g. a square inch, a square rod, etc. There are exceptions to this, as the acre used in land measurement which consists of 160 square rods. 356. AREA. The ratio of a given surface to the unit of surface is the area of the surface. It is expressed in terms of the unit used, e. g. 15 square inches, 24 square yards, etc. When thus expressed area tells two things ; first, the number of times the unit is contained in the surface and second, the name of the unit used. PROPOSITION I. 357. THEOREM. Two rectangles having equal bases are proportional to their altitudes. Given two* rectangles P and P' with equal bases m and altitudes n and n' respectively. 170 PLANE GEOMETRY To Prove P -=^ P' ri The altitudes n and n' are assumed to have a common unit of measure. The case in which they do not, i. e., in which they are incommensurable, will be considered later. SUG. 1. Suppose n contains this common unit k times and that n' contains it k' times. What then is the ratio of n to n' ? Auth. ? 2. Through the points of division in n and n' draw lines parallel to the bases. The fig- ures thus formed are rectangles. Why ? They are congruent. Why ? Therefore one of them can be used as a unit of measure. 3. How many times is this unit rec- tangle contained in P ? In P' ? What is the ratio of P to P' ? 4. Compare the ratio of the altitudes with that of the rectangle. Auth.? Therefore 358. COR. // two rectangles have equal altitudes they are proportional to their bases. SUG. Either side of a rectangle may be re- garded as the base. 359. COR. // two rectangles have a common dimen- sion, they are proportional to the other dimensions. The dimensions of a parallelogram or a triangle are its base and altitude. AREA OF POLYGONS 171 PKOPOSITION II. 360. THEOREM. Tico rectangles are propor- tional to the products of their dimensions. Given rectangles M and M' with dimensions a, b and a', b' respectively. M axb To Prove = a'xfc' Proof. SUG. 1. Construct a rectangle P with one dimension of M, say a, and one dimension of IT, say 6'. 2. What is the ratio of M to P ? Of P to J/' ? Why? 3. Find the product of these two ratios. What then is the ratio of M to M ' ? Therefore PROPOSITION III. 361. THEOREM. The area of a rectangle equal* the product of its base and its altitude. Given rectangle P with dimensions m and n and unit P'. To Prove area P = m x n times P'. 172 PLANE GEOMETRY . SUG. The ratio - = = HIM. Why? P' Ixl ' P = mnP' or area P mn of the units of area. The theorem is here stated in the usual abbreviated form. Literally interpreted it would imply that area is a number and that the "base ana altitude" also are numbers. Stated in full and as it should be interpreted, it would read: The area of a rect- angle equals the product of the measures of the base and alti- tude times the unit of measure. All the theorems of areas will be stated in the abbreviated form. The name of a polygon is frequently used to designate its area. Polygons and circles are the surfaces to be measured in plane geometry. 362. If the sides of the unit of measure are exact divisors of the corresponding sides of the rectangle a simple means of determining the area is to lay off the unit upon the rectangle, noting that the number of times the base of the unit is contained in the base of the rectangle is the number of area units in one row and that the number of times the altitude of the unit is con- tained in the altitude of the rectangle is the number of rows. Then by arithmetical analysis it is seen that the number of the rows times the number of the units in each row is the total number of units of area. The demonstration of 357 covers all cases of practical value for if the altitudes are incommensurable an approximate unit of altitude can be taken as small as the needs of the given case re- quire. For instance, if the altitudes are 3 in. and VlTin. the ratio > of the altitudes will be approximately if 1 in. be used r.s *)/\ ^nn the unit; if the unit is .1 in.; if the unit i .01 in.; etc. The answer in this last case is correct to less than the area of a strip the length of the rectangle and .01 of the linear unit in width. Thus the accuracy of the area computed will depend AREA OF POLYGONS 173 upon the accuracy of the measurement of the dimensions, for the demonstration by 357 can be carried rigorously to an approxi- mation far beyond the skill of man to measure in any given case. All actual measurement gives but an approximation of the true value of the magnitude and the closeness of the approximation in measurement is usually determined by the degree of accuracy required. PROPOSITION IV. 363. THEOREM. The area of a parallelogram equals the product of its base and altitude. Given Parallelogram DEFG with base b and alti- tude a. To Prove area of DF = ab. Proof. SUG. 1. From two adjacent vertices as D and E, drop perpendiculars to the opposite side, as DM and EN. 2. Compare A DMG with A EXF. 3. Compare the areas of the parallelo- gram DF and the rectangle DN. 4. What is the area of DF + A P = A tf . Is this relation still true if for N, D, P we substitute any simi- lar polygons? 10. The square on the side of a triangle opposite an obtuse angle equals the sum of the squares upon the two other sides plus twice the rectangle formed by one side and the projection of the other side upon that side. Givfix A ^BC with /.B obtuse, AE the sq. opposite L B, with BM and BE the squares on the two other sides, BB' the projection of AB upon BC, and B'F the rectangle. To PROVE AE = BM -f BH + 2 B' F. N D' 3T? M Sug. 1. Assume the theorem in its algebraic form from 338. Substitute for the algebraic products the areas which they represent in the above figure. (Draw BE, CD and extend AB to D'C.) SUG. 2. From each vertex of the given triangle drop perpendiculars to the farther side of the opposite squares, or to these sides extended if need be. Connect the vertices of the triangle with the opposite vertices of their respective opposite squares. Work out a demonstration similar to 376. ] 1 . The square upon the side opposite an acute angle of a triangle equals the sum of the squares upon the other two sides minus twice the rectangle formed by one of those sides and the projection of the other upon it. GIVEN A ABC with sides a, b, c. To PROVE the relation O2z=&2-f-c2 2bm, in which 0,2, faj c* 186 PLANE GEOMETRY represent squares on the sides a, fe, c, respectively and fern the rectangle formed by side fe and the projection of c on &. SUG. 1. Prove from 376 as in preceding exercise. Sue. 2. Draw the auxiliary lines and prove geometric- ally as in the preceding exercise. 12. What is the ratio of &MNO to A PMO, ZM being 90? Fig. 1. 13. Prove geometrically (a + 6)2 = a 2 + 2afe + 62. Fig. 2. 14. Prove geometrically (a fe) 2 = a 2 2afe + & 2 . Fig. 3. 15. Prove geometrically (a + fe) (a fe) a- fe 2 . Fig. 4. a b fl-6 b ... r/ " "' i b *! 1 f -b Fig. 1. Fig. 2. Fig. 3. Fig. 4. 16. Prove by a figure the following formulas: a (a fe) 2 -ab a (fe + c) = afe + ac (a + fe) (a + c) = a 2 + a & + ac -f fe c . (ct -f- &) (a c)izr2-|-a6 ac fee (a &) (a c) = a2 a& ac + fee. (a-f fe) 3 -f (a 6) 2 = 2:i a + 2fe 2 (a + 6) (c + d) = ac + ad + fee + fed. 1 7. To construct a square equal to the sum of two given squares. GIVEN a square on a and a square on fe. Fig. 5. To CONSTRUCT a square equal to #2 -f fe2. SUG. Construct a right triangle of which a and fe are the legs. Complete construction. The proof is left to the pupil. 18. Construct a square equal to the sum of three given squares; of four given squares; of n given squares. 19. Construct a square equal to the difference between two given squares. PKOBLEMS OF CONSTRUCTION 187 PROPOSITION XIV. 379. PROBLEM. Upon a given base to construct a rectangle equal to a given rectangle. Given rectangle DF with adjacent sides b and c and sect a. To Construct a rectangle on a equal to DF. SUG. 1. Let x represent the altitude of the required rectangle. Then b X c = a X x. 2. From this derive a proportion with x as the fourth and unknown term. 3. Find x and complete the construction. 4. Show that the constructed rectangle equals DF. PROPOSITION XV. 380. PROBLEM. Upon a given sect as base, to con- struct a rectangle equal to a given square. Given sect b and a square on sect a. To Construct on b a rectangle equal to the given square. SUG. 1. Let x be the altitude of the required rectangle. Then a 2 = b X x. 2. From this derive a proportion with x as the fourth and unknown term. What prob- lem is involved in the finding of xf 3. Complete the construction and prove the constructed rectangle equal to the given square. 188 PLANE GEOMETRY 1. Upon a given sect as base construct a rectangle the area of which shall equal the sum of the areas of a given square, a given trapezoid, and a given triangle. 2. The area of a square inscribed in a circle is one-half the area of any square circumscribed about the circle. 3. Construct a parallelogram with a given base and a given angle, the area of which shall equal that of a given rectangle. 4. Find the dimensions of a rectangle the perimeter of which is 26 in. with an area of 40 sq. in. PROPOSITION XVI. 381. PROBLEM. To construct a square equal to a given rectangle. . -I > I Given a rectangle with adjacent sides a and b. To Construct a square equal to rectangle ab. SUG. 1. Let x represent the side of the re- quired square. Then x 2 = ab. 2. Derive a proportion and find x. 3. Complete the construction. 5. Verify the constructions in 379, 380, and 381 by meas- uring the dimensions and computing the areas. 6. Construct a square equal to a given trapezoid. 7. Construct a square equal to a given triangle. 8. Construct a square equal to a given parallelogram. 9. If the hypotenuse of a right triangle is 15 ft. and the ratio of its legs is what is its area? 10. Cut off one-third of a parallelogram by a line through a vertex. Cut off one-fourth in the same manner. 11. Construct an isosceles triangle having a given altitude and equal to a given triangle. 12. Bisect a parallelogram by a line parallel to a given line. 13. Construct a triangle equal to a given triangle, M, on base, o, and with its opposite vertex in a given line, x. Can this line be parallel to the base? EXERCISES 189 14. The diagonals of two squares are 5 ft. and 9 ft. respect- ively. What is the diagonal of a square equal to their sum? 15. Divide a triangle by a line through trisection points on one side into parts proportional to 1, 2, 3. 16. The side of an equilateral triangle is 10. What is tbi length of a side of a regular hexagon of the same area? Of a square of the same area? 17. The diagonal of a rectangle is 10 and the ratio of the sides is f. Find the area. If the diagonal is 40 and the ratio of the sides is J., find the area. 18. The difference between the squares of two sides of a tri- angle equals the difference between the squares of the projections of those sides upon the third side. 19. If two circles are tangent, internally or externally, chords of the one passing through the point of contact are divided pro- portionally by the other. 20. Draw a chord through a given point within a circle which shall be divided by this point in the ratio 1 Also the ratio TO Sue. Use preceding ex. 21. Draw a secant from a point without a circle terminated by the circle such that its ratio to its external segment s.hall be 3 to 1. Also . Is this problem ever impossible? 22. Through a given point draw a line so that its distances from two given points shall be in a given ratio. 23. What is the locus of points which divide all chords of a circle passing through a given point, internally or externally, in a given ratio? 24. Let DEFG ... be any polygon. Jo 4 n the vertices to a point O. Through D' t any point on OD, draw a line parallel to DE terminating in E' on DE. Through E' draw a line parallel to EF, etc. Prove the polygon D' E' F' G' ... similar to DEFG. . . 190 PLANE GEOMETRY Two similar polygons can always be placed in such a position with respect to some fixed point. This property is characteristic and is sometimes taken as the definition of similarity^ The point in which the lines joining homologous vertices meet is called.' the center of similitude. 24. The area jf a triangle equals one half the product of two sides and the sine of the included angle. This proof covers only the case in which the included angle is acute. The general theorem is proved in trigonometry. GIVEN A ABC with sides a and c including the acute angle B. To PROVE area ABC = % ac sin B. PROOF. Area ABC - \ ali & and h & = c sin B. Hence the theorem. Fig. 369. 25. Drive three stakes in the School grounds so as to form an acute triangle. Measure two sides and the included angle and compute the area by ex. 24, also measure a base and altitude and compute and compare results, Review your work if not approx imately the same. CHAPTER V. REGULAR POLYGONS. 382. REGULAR POLYGON. A polygon which is both equilateral and equiangular is regular. The equilateral triangle and the square are regular polygons. PROPOSITION I. 383. THEOREM. An equilateral polygon in- scribed in a circle is a regular polygon. Given AD, an equilateral polygon inscribed in a circle. To Prove that AD is a regular polygon. Proof. Sue. 1. According to the definition, 382, what remains to be proved in order that AD be regular ? 2. By what may the angles A, B, C, etc., be measured? 3. How do they compare with each other? 4. Test AD by the definition of a regular polygon. Therefore 384. COR. // a circle be divided into any number of equal parts, the lines joining the points of division form a regular polygon. Proof left to the pupil. 192 PLANE GEOMETRY PROPOSITION II. 385. THEOREM. A circle can be circumscribed about a regular polygon. A circle can be inscribed in a regular polygon. Given a regular polygon, AD. To Prove. I. A circle can be circumscribed about AD. SUG. 1. At M and N the mid points of two adjacent sides erect perpendiculars extended un- til they meet as at 0. Why do they meet? 2. Join to the vertices A, B, C, etc. Compare 4 AOE and BOC ; L 1, 2 and 3. 3. A DOC = A COB. Why? Hence OD = OB. Why ? 4. Similarly all the sects OA, OB, OC, OD, etc., are equal and hence a circle with center can be circumscribed. Why? To Prove. II. SUG. 1. What was proved concern- ing the successive A AOB, BOC, etc.? 2. Compare the sects OM, ON, etc. Auth. 3. Complete the demonstra- tion. Therefore Query. How many sides in the polygon .1 /> ? See the theorem. REGULAR POLYGONS 193 386. RADIUS OF A REGULAR POLYGON. The radius of the circumscribed circle is the radius of a regular poly- gon. 387. APOTHEM OF A REGULAR POLYGON. The radius of the inscribed circle is the apothem of a regular poly- gon. 388. CENTER OF A REGULAR POLYGON. The center of the inscribed and circumscribed circles is the center of a regular polygon. 389. ANGLE AT THE CENTER OF A REGULAR POLYGON. The angle formed by two radii drawn to two adjacent vertices of the polygon is the angle at the center of a regular polygon. From the definitions just given the following corol- laries can be deduced. The pupil should prove them. 390. COR. 1. The angle at the center of a regular polygon is equal to four right angles divided by the number of sides of the polygon. 391. COR. II. An interior angle of a regular poly- gon is equal to the sum of all the interior angles of the polygon divided by the number of sides of the polygon. 392. COR. III. The angle at the center of a regular polygon equals an exterior angle and is the supplement of an interior angle of the polygon. 393. COR. IV. The radius of a regular polygon bi- sects the angle of the polygon to which it is drawn. 1. A farm is 320 rods long, and rectangular in shape. From one end a square farm is cut off, leaving 120 acres. How wide is the farm and how many acres in the entire piece? 2. From any point M in side BC of A ABC draw c, sect meet- ing AB produced in D, so that MD is bisected by AC. Also, so that MD is divided by AC in any given ratio. 194 PLANE GEOMETRY PROPOSITION III. 394. THEOREM. // a circle is divided into any number of equal parts, the tangents drawn through the points of division form a regular cir- cumscribed polygon. Given the circle ABC . . . . , divided into equal parts AB, CD, EF. . ., with tangents at the points A, B, C,. . . forming the polygon A 'B '' . . . . To Prove that A'B'C' ... is regular. SUG. 1. A A' AB, B'BC, etc., are isosceles and congruent. Why ? 2. Compare AA' , A'B, BB' , B'C, etc. 3. Prove A'B' = B'C' = C'D', etc. 4. Prove A. A' , B' , C', etc., equal. 5. Apply the definition of a regular polygon. Therefore 395. COB. I. // the vertices of a regular inscribed polygon are connected with the mid points of the arcs subtended by the sides, a regu- lar inscribed polygon of double the number of sides is formed. 394. 396. COB. II. The perimeter and the area of a regu- REGULAR POLYGONS 195 lar inscribed polygon are less than the perimeter and area respectively of a regular inscribed polygon of dou- ble the number of sides. 397. COR. III. // a regular polygon is circum- scribed about a circle and tangents are drawn at the mid points of the intercepted arcs, a regular circum- scribed polygon of double the number of sides is formed. 394. 398. COR. IV. The perimeter and the area of a regular circumscribed polygon are greater than the peri- meter and the area respectively of a regular circum- scribed polygon of double the number of sides. 1. The ratio of similitude of two similar polygons is A and the sum of their areas is 518 sq. in. Find the area of each. 2. Find the base of a rectangle with an area of 108 sq. ft. and an altitude of 6 ft. Compare the method with that of 379. 3. Find the area of a right triangle with an hypotenuse of 1 ft. 8 in. and one leg of 1 ft. in length. 4. The area of a circumscribed polygon equals one-half the product of its perimeter and the radius of the circle. 5. If the mid points of two adjacent sides of a parallelogram be joined the area of the triangle thus formed is one-eighth that of the parallelogram. PROPOSITION IV. 399. THEOREM. // a regular polygon is in- scribed in a circle and tangents are drawn at the mid points of the arcs subtended by the sides: I. A regular circumscribed polygon is formed. II. The sides of the circumscribed polygon are parellel to the sides of the inscribed polygon, each to each. 196 PLANE GEOMETRY i III. The vertices of the circumscribed polygon lie in the extended radii of the inscribed polygon. A' C' Given a regular inscribed polygon ABC , with M f N, P, . . . . the mid points of the arcs subtended by the sides AB, BC, CD, and tangents A'B', B'C', C'D', through the points M, N, P, To Prove. I. A'B'C' . . . is regular; II. AB I! A'B', etc.; III. A' lies in OA produced, etc. I. SUG. 394. II. SUG. Draw radius OH 1 AB and ex- tend it. "Where does it meet arc AB*t How do AB and A'B' lie with reference to OH! Com- plete the argument. III. SUG. 1. LA=LA'. Why? 2. OA bisects Z A and OA' bi- sects Z A'. Why? 393. 3. OEM is a straight line LAB and A'B'. Hence Z MOA = Z MOA' . Why? 4. Hence A' lies on OA. Why? Therefore PROPOSITION V. 400. PROBLEM. To inscribe a regular hexagon in a circle. Given a circle with radius r. To Inscribe a regular hexagon. REGULAR POLYGONS 197 Solution. SUG. 1. The radius and side of the re- quired hexagon are equal. Why ? 2. With the dividers lay off six equal arcs. What is the length of the subtended chord ? 3. Complete the constructior PROPOSITION VI. 401. THEOREM. // the radius of a circle is di- vided into extreme and mean ratio, the greater segment equals one side of a regular decagon in- scribed in the circle. Given O with radius A divided at C into extreme and mean ratio, OC being the greater segment. To Prove OC equal to one side of a V / regular decagon inscrioed in the circle. 1. Draw the chord AB equal to OC and join and C to B. 2. It is to be shown that arc AB is one- tenth of the circle. 3. - = . Why? 327. OC AC OA AB 4. - -- WViv^ AB AC * 5. A OAB <-* A BAG. Why ? 283. 6. Compare Z ABC with Z 0; L CBO with Z 0; L ABO with Z 0. 7. Hence Z is what part of 2 rt. A. ? Of 4 rt. 1 ? 8. Hence AB subtends what fractional part of the circle ? Therefore 198 PLANE GEOMETRY 402. COR. A regular decagon can be inscribed in a circle by dividing the radius in extreme and mean ratio and taking the greater segment as the side of the required polygon. 1. Inscribe a regular polygon of 20 sides. J2. Inscribe a regular pentagon. 3. Inscribe a regular polygon of fifteen sides, called a penta decagon. Sue. 1. Draw a chord AB equal to the radius, and AC equal to the side of a regular inscribed decagon. 2. What part of the circle is subtended by AB? by AC? 3. Hence what part of the circle is subtended by CB? 4. Complete the construction. 4. What regular polygons can be inscribed the construction of which can be based upon that of the regular penta-decagon? 5. If the area of a regular inscribed triangle is 30, what is the area of the regular circumscribed triangle? 0. If the area of an inscribed square is 45, what is the area of the circumscribed square? 7. What is the numerical ratio of the apothem of a regular inscribed hexagon to that of a regular circumscribed hexagon? PROPOSITION VII. 403. PROBLEM. To inscribe a square in a circle. Solution. SUG. 1. Two vertices must be the ex- tremities of a diameter. Why? 2. How can the two other vertices be located? 3. Draw a diameter and complete the construction. Proof. 8. Inscribe a regular polygon of 8 sides; of 16 sides; of 32 sides. R KGTLAR POLYGONS 199 PROPOSITION VIII. 404. THEOREM. Regular polygons of the same number of sides are similar. A' B' V- D' Given AD and A'D' , two regular polygons of the same number of sides. To Prove AD ^ A'D'. Proof. SUG. 1. What must be established to prove the similarity? 2. Compare the corresponding angles A and A' , B and B' , etc. 3 = 1 '= 1 ' Bte - Why? 4. Compare the ratio AB A'B' AB BC - with - : - with -- . BC B'C' A'B' B'C' 5. Compare BC . B'C' BC .;, CD with - : -- witu -- etc. CD C'D' B'C f C'D' AB BC CD 6. 2-== -- --- etc. A'B' B'C' C'D' Therefore 1. What is the locus of the centers of circles which are tan- cre::'t to a given line at a given point? 2. Two parallel chords of a circle are respectively 36 and 48 inches long. The radius is 30 inches. Find the distance between the chords. 200 PLANE GEOMETRY 1. Find the dimensions of a rectangle which has a perimeter of 16 in. and an area of 15 sq. in. ; the area of one with a perim- eter of 28 ft. and an area of 10 ft. 2. If a parallelogram be inscribed in or circumscribed about a circle, the diagonals pass through the center. PROPOSITION IX. 405. THEOREM. The perimeters of two similar regular polygons have the same ratio as their radii and their apothems. \ __/ Given AD and A'D' } two similar regular polygons with OA and 0' A' their respective radii, OM and O'M' their respective apothems, AB and A 'B' two homolo- gous sides, and p and p' their respective perimeters. p OA OM To Prove = --- = -- . p' O'A' O f M f Proof. SUG. 1. Compare ; with - . Autli. p' A' B' 2. Draw OB and 'B' Then A AOB = A A'O'B'. Why? OA OM AB 4. Complete the proof. Therefore 406. COB. The areas of two similar regular poly- gons have the same ratio as the squares of their radii and the squares of iheir apothems. The proof similar to that of 405 is left to the pupil. REGULAR POLYGONS 201 PROPOSITION X. 407. THEOKEM. The area of a regular polygon is equal to one-half the product of its perimeter and apothem. Given a regular polygon AD, with area denoted by K, perimeter by p, and apothem by a. To Prove K = - p x a . & Proof. SUG. 1. What is the area of A AOE1 Of BOC1 2. What is the area of each A ? 3. What is the sum of the areas of all the A? Therefore In answering Sug. 3, let JL a be the unit of addition and AB, BC, etc., be the respective coefficients; or indicate the addi- tion and divide by the common factor la. 408. POSTULATE. // the number of sides of a reg- ular inscribed polygon be increased indefinitely, th# apothem approaches indefinitely near the radius in length, likewise the perimeter approaches indefinitely near the circle and the area of the polygon approaches indefinitely near the area of the circle. A more exact statement of the postulate is as follows : By sufficiently increasing the number of sides of a regular inscribed polygon the difference betic&en the apothem and the radius can be made less than any given 202 PLANE GEOMETRY number, however small. The same is true of the perim- eter of the polygon and the circle and the areas of the two figures. 1. What must be the nature of a parallelogram in order that a circle can be circumscribed about it? 2. [f the radius of a circle is 12 inches, what is the length of a side of an inscribed square? What is the length of the apothem? What is the area of the square? 3. Find the area of the square in the preceding example by use of the apothem as a check upon the first computation. . If A and A ' are two similar polygons, m and m' two homologous sides, and A = 2A' f then m m' V 2. 5. If A' is 16, find A. If A' is 12, find A. If A is 16,find A'. If A is 12, find A 1 . Find the coefficient of W if a is i a' if a is a' ; if a is 3a' ; if a is KA. 6. If one acute angle of a right triangle is 60, prove that the area of the equilateral triangle constructed on the hypotenuse is equal to the area of a rectangle the adjacent sides of which are the two legs of the right triangle. 7. If two triangles have two sides of one equal to two sides of the other respectively and the included angles supplementary, they are equal. A , The diagonals of a parallelogram di- vide it into four triangles equal in area. / / 9. Draw two concentric regular hexagons, one being twice the size of the other. SUG. Having drawn one, with radius OA, let x = OA' be the unknown radius of the other. Then x can be found from the proportion A AOB = 1 = . A A'O'B' x s 10. If 10 is the length of the radius of a regular hexagon, what is the length of the radius of a regular hexagon twice as large? 11. Divide a regular hexayou into four equal parts by con- centric regular hexngons. REGULAR POLYGONS 203 PROPOSITION XI. 409. THEOEEM. Tico circles have the same ratio as their radii. Given two circles, c and c' with radii r and r', respect- ively, p and p' the respective perimeters of the similar inscribed polygons, P 2 , p/ being the respective perime- ters of two similar inscribed polygons of double the num- ber of sides, etc. To Prove c' r' Proof. 1. By 405 '?*- = *- = *- ...... = *L. Pi P a ' P* ?' 2. According to 408 the difference between the circles and the perimeters of the inscribed polygons can be made as small as one chooses. Hence if x and x' be any small numbers there must be a point in the sequence of inscribed poly- gons where c p and c' p' are less than x and x' respectively. If these small differences be denoted by y and y f , then c p y and c' p' ~y'- r> v r 3 Hence may be written r r' c-j_ __ c'y' r r' 4. From 3 follows - = --- & r r r r 5. But since y and y' can be taken smaller than any given number the expression in paren- thesis may for all practicable purposes be neg- lected, /> / 6. Hence Therefore r r 204 PLANE GEOMETEY < 410. COR. I. The ratio of a circle to its diameter is a fixed number whatever the radius. f* * O-* // Proof. From the above proposition -=-== c r 2r' d' in which d and d' are the respective diameters. From XI y-,' this follows - = . From this proportion follows the d d' corollary. This ratio is represented by the Greek letter TT. /> 411. COR. II. - = TT or c = Trd = 2?rr. d 412. COR. III. The area of two circles have the same ratio as the squares of their radii. Let C and C f represent the two areas. Complete the proof in a manner similar to that of 409. In this proof the area of a circle can be approximated to any desired degree of accuracy, far beyond any de- mands of practical measurement. 413. COR. IV. The areas of two circles have the same ratio as Uie squares of their diameters. 414. COR. Y. Areas of similar sectors of two circles have the same ratio as the squares of their respective radii and diameters. 1. Two tangents are drawn to a circle of 8 in. radius from a point 12 in. from the center. Find the length of the chord join- ing the points of tangency. 2. The legs of a trapezoid are eac.h 15 inches and the bases are 12 and 30 inches respectively. Find the area. 3. Two registers similar in form are respectively 10 and 20 inches in width. What is the ratio of the amounts of air pass- ing through them, the pressure being the same? 4. Two circular windows are respectively 24 and 30 inches in diameter. What is the ratio of their respective efficiencies for admitting light? AREA OF CIRCLES 205 1. What is the locus of the vertices of all isosceles triangles on a given base? 2. Given A ABC with D any point in BC extended. Find a point E in AB or in AB produced such that the area of A EBD will equal that of &ABC-. will be one-half the area of &ABC: will be double the area of A ABC : will have any given ratio to the area of A ABC. PROPOSITION XII. 415. THEOKEM. The area of a circle is equal to one-half the product of its length and its radius. Given a circle o, with radius r, length c and area A. To Prove A = ^ cr. Proof. SUG. 1. If a sequence of regular polygons be inscribed, in each succeeding polygon of the sequence the number of sides being doubled, the apothem approaches the radius, the perimeter ap- proaches the circle, and the area of the polygon approaches that of the circle. 408. 2. The area of any of the polygons is J ap, a and p representing the apothem and perim- eter respectively. 3. Substituting for these quantities those to which they, by this process, are made to approach, one obtains A = J cr. Therefore NOTE. A more rigorous demonstration of the above theorem and of similar theorems which follow can be made by use of the theory of limits. In such a demon- 206 PLANE GEOMETRY * stration the possibility of making the substitution in step 3 of the above demonstration would be considered and proved in some detail. It is not thought wise to in- troduce such considerations at this point. 416. COR. I. The area of a circle equals nr 2 or 4 Proof. Substitute for c, in -J cr, its value 2 TT r. 417. COB. II. The area of a sector equals one-half the product of its radius and its arc. Proof. SUG. 1. Prove by a method similar to that of 297 that two sectors are to each other as their angles. 2. Sector of arc a a Sector of arc 2-n-r 2-n-r. 3. But a sector of arc 2?rr has an area of Trr 2 . Substitute this and obtain a sector of arc a = | a X r. NOTE. Since the formulas for area involve the number TT, their accuracy, as well as that of the formula for length, depend upon the accuracy with which the number TT is computed. The pupil should note that this computation has not yet been made. 419. 1. The angle of a sector is 30 and the radius is 24 ft. What is the area of the sector? Express in terms of TT. 2. The area of a sector is 88 sq. in. and its angle is 60. Find the diameter of the circle. 3. The moon has approximately one-fourth the diameter of the earth. At a point equidistant from them what is the ratio of their respective "moonshine" powers? 4. Two rectangular windows similar in form are respectively 2' 6" and 3' in width. What is the ratio of the amounts of light they admit, other conditions being equal? 5. What is the locus of a point equidistant from two con- centric circles? 6. Cut the sides of an equilateral triangle with three sects so as to form a regular hexagon. AREA OF CIRCLES 207 PROPOSITION XIII. 418. PROBLEM. Given the radius of a circle and the side of a regular inscribed polygon, re- quired to find the side of a regular inscribed poly- gon of double the number of sides, in terms of the given quantities. Given a circle with radius r, AB a side of a regular inscribed polygon, and AC a side of a regular inscribed polygon of double the number of sides. To Determine A C in terms of AB and r. SUG. I. In A APC express AC in 'terms of AP and CP ( 376) and then in terms of AB and CP. 2. Express CP in terms of r (i. e. OC) and OP. Express OP in terms of r (i. e.AO)and AP ( 376) and then in terms of r and AB. 3. Express CP in terms of r and AB. 4. Express AC in terms of r and AB. This relation when simplified becomes AC = 5. If r be taken equal to unity, this be- comes AC = ^2 - V4-Z0 2 A partial statement of the details of the above steps is as follows : AC = 208 PLANE GEOMETRY 4 By means of this formula, if the length of the perim- eter of any regular inscribed polygon is known, the length of the perimeter of a regular inscribed polygon of double the number of sides can be computed. From this result the perimeter can be computed in like manner for a polygon the number of sides of which is again dou- bled. This process can be continued indefinitely. PROPOSITION XIV. 419. PROBLEM. To compute approximately the ratio of a circle to its diameter. SUG. 1. For convenience the radius is taken as unity. Why may this be done ? 2. The perimeter of a regular inscribed hexagon is 6 which may be regarded as a first ap- proximate value of the circumference. The first t) 6 approximation of TT is then v*= 3. From the formula of 418, the value of s l2 , which represents the length of a side of a regular inscribed polygon of 12 sides is 2- V (4-12; = . 51763809 and consequently p 12 , or the perimeter, is 12s 12 , or 6.21165708. From this is obtained a second and closer approximation of AREA OF CIRCLES 209 TT, . e - = 6.21165708 = 3.10582854. d 4. Similarly - (5.5176380)* and p 24 24 x s 24 . 5. Certain of the computed results are given in the following table : No. Sid's One Side Perimeter 7T C 12 24 48 96 J92 1 V2-V3~ = .51763809 6. 6.21165708 6.26525722 6.27870041 6.28206396 6.28290510 3 3.10582854 3.13262861 3.13935020 3.14103198 3.14145255 V2 V4 (.51763809) 2 = .26105238 ^ 1 V4 (.26105238) 8 = .13080626 ^ 2 V 4 (.13080626) 2 = .06543817 V 2 V 4 (.0654381 7) 8 =--- .03272346 The approximation of TT in common use is 3.1416. For ordinary work 3^ is sufficiently accurate. 1. Inscribe a regular triangle. A regular polygon of twelve sides. Of twenty-four sides. 2. Measure the circumferences of several circular objects as a plate, the end of a pail, barrel, or stove pipe. Divide the cir- cumference by the diameter. Average the several results to deter- mine an approximate value of TT. 3. Compute the length of the perimeter of a 24 sided poly- gon, P 24 , with radius unity and compare the result with that given in the table of 419. 4 If the radius is unity, what is the length of one side of a regular inscribed triangle? 5. The radius is one and a side of a regular inscribed tri- angle is V~3~. Use the following formula to find the side of a 210 PLANE GEOMETRY regular inscribed hexagon. Verify the result by 4!,S. Formula. *.= V 2-V (4-VJ .* . . A second method of solving the problem of M 418 is as follows: /TN To find AC in terms of r and AB. ( P J A - Sue. AC=-^MC X PC. ic 420. SCHOLIUM. Archimedes (born 287 B. C.) found an approximate value for TT. He proved that its value is between 3^ and 3-^f . In modern times the value of TT has been computed to a large number of decimal places, two men, Clausen and Dase independently of each other having computed the value to the two-hun- dredth decimal place. Other computers have given the value to over five hundred decimal places but their re- sults have not been verified. The number TT is incom- mensurable with 1, i. e. it is neither an integer or a fraction, and hence cannot be expressed exactly by any number of decimal places. 1. The radius of a circle is 15 rods. Find its length, or circumference, and its area. 2^ With a tape line measure the circumference of a tree and compute its diameter. 3. Draw a circle on paper or blackboard. Measure its diam- eter and compute its length and its area. 4. A circular silo is 30 ft. in diameter. How many squaro feet in its floor? 5. Given a circle with radius 5. What is its circumference or length and area? 6. The length of a circle is fourteen. What is its diameter and its area? 7. The area of a circle is 28. Find the length and the diam- eter. 8. Four 6 in. circles are tangent to each other. What is the area of the smallest square that can enclose thrni? Auth. EXERCISES 211 9. "What is the area of the space enclosed by the circles of the preceding exercise? 10. What is the area of the circle externally tangent to the above four circles? 11. What is the area of the circle internally tangent to the above four circles. 12. Three 6 in. circles are tangent externally, what is the area of the triangle of their centers? What is the area of the space included between the circles? 13. What is the length of the apothem and what is the area of the triangle circumscribing them? 14. What is the area of the circle tangent to them inter- nally? 15. If the area included by three equal circles tangent to each other is 5 acres, what is their radius? 16. Within a given circle construct 3 equal circles tangent to each other and to the given circle. Ex. 2, p. 37. 17. Construct three circles tangent to each other and exter- nally tangent to a given circle. See ex. 2, p. 37. 18. Within a regular triangle construct three equal circle?, each tangent to the two others and to two sides of the triangle. 19. A circle is escribed to a triangle when it is tangent to one side and to the two other sides produced. A triangle may have three escribed circles. Given a triangle, construct the three escribed circles. Sue. Consider the principle of inscribing a circle in a triangle. 20. Find the area of the ring included between two concen- tric circles with radii of 8 and 10 inches respectively. 21. Find the length of a sect which is a chord of one of two concentric circles and a tangent to the other. Express the length in terms of the two radii. 22. Circles are described upon the three sides of a right tri- angle as diameters. Show that the one on the hypotenuse equals the sum of the others. 23. The diameter of a circle is one of the legs of an isosceles triangle. Show that the base is bisected by the circle. 212 PLANE GEOMETRY . 24. A three inch and a five inch drain tile unite. What size of tile is necessary to continue the drain? 25. Three 5-in. tiles unite. What size of tile should be used at the union? Two methods. 26. If a straight line cuts two concentric circles, the segments included between the circles are equal. 27. If two diameters intercept at right angles, the sum of the squares upon the segments equals the square upon the diameter. 28. A saw log is 15 in. in diameter. What is the area of the cross section of the largest squared timber which can be sawn from this log? What proportion of the log goes into slabs? 29. If the above log were cut into a timber with rectangular cross section eight inches thick, what would be the cross section area? If the slabs are wasted, which of these two methods of sawing is more profitable? 30. The radius of a circle inscribed in a regular triangle is one-third its altitude. 31. Show that seven equal circles can be so drawn that one of them is " internally " tangent to the six others and that each of these six is tangent to two of the six. 32. Show that one circle can be so drawn as to include the circles of the preceding exercise and be tangent to six of them. 33. What part of the area of the large circle in above ex. is included in the area of the seven equal circles? 34. If the radius of the equal circles is 4, what is the area of that portion of the large circle which is external to the small circles ? 35. Six dimes can be placed so that each is tangent to two others and also to a seventh dime. What is the ratio of the diam- eter of the dime to that of the circle which would circumscribe them all? 36. The sides of two regular hexagons are 3 and 5. If the area of the former is M, what is the area of the latter? 37. A regular hexagon of 63 square ft. is inscribed in a cir- cle. Another is circumscribed about the circle. What is the area of the latter? CHAPTER VI. INCOMMENSURABLE MAGNITUDES. 421. In the discussion of quantities thus far only commensurable magnitudes have been considered, ex- cept in theorems involving TT. INCOMMENSURABLE MAGNITUDES. (249.) Two mag- nitudes of the same kind that can not both be exactly measured by the same unit however small are incom- mensurable. Thus V 3 and 5 are incommensurable. 422. INCOMMENSURABLE RATIO. The ratio of two in- commensurable quantities is an incommensurable ratio or an incommensurable number. Thus the ratio y3 is an incommensurable number. Approximations to the value of an incommensurable number may tag obtained by neglecting fractional parts of the unit used. For example, the use of 1 as a unit gives as a first approximation of ^ 3 the number 1. The use of .1 as a unit gives 17. The use of .01 as a unit gives 173. The use of .00001 as a unit gives 173205. This approximation may be carried to any desired degree of accuracy by the use of still smaller units. An incommensurable number is neither a whole num- ber nor a fraction, for if it were either the original quan tities would not be incommensurable with respect to each other. The pupil wust not lose sight of the fact that an incom- mensurable number is as truly existent as any other. This fact 214 PLANE GEOMETRY . is well illustrated by the consideration, for example, of the diag- onal of a 1 in. square. That such a sect exists and that it has a length is not questioned. Geometry proves that its length expressed in inches is V~2T This is an incommensurable number for if not it is an integer or a fraction. But it is neither of these because it is a number that when squared gives 2. It is quite evident that no integer when squared gives 2 and also that no fraction when squared will give an integer. OL 2 Thus if = , a and & cannot be incommensurable with o 3 respect to each other. This is seen as follows: By proportion ^~ . Put & a = r and obtain a = 2r and from this a 2 The results in all practical measurements are but ap- proximations of the magnitudes measured, depending for accuracy upon the skill of the operator and the pre- cision of the instruments used. But when the results obtained are within the degree of accuracy required, they are used as if they were the actual magnitudes and not approximations. Hence the previous demonstrations involving incom- mensurable magnitudes satisfy all demands of practical usage. For theoretical purposes it is interesting and im- portant to know the absolute truths in the cases before considered and the discussion of such requires addi- tional demonstrations of considerable rigor. 423. CONSTANT. A quantity that is given the same value through a discussion is a constant. 424. VARIABLE. A quantity which is given different values in a discussion is a variable. In general varia- bles in changing values pass through a succession of val- ues according to some law which serves to distinguish them from other variables. INCOMMENSURABLE MAGNITUDES 215 425. LIMIT OF A VARIABLE. When a variable by its law of change differs from a constant by a variable quan- tity which may become and remain less than any as- signed quantity however small, but not zero, this con- stant is the limit of the variable. A variable may or may not attain its limit. This is a matter dependent upon the law under which the varia- ble exists. The successive approximations of V 2 may be considered as the successive values of a variable which has V 2 as its limit. In this case the variable does not attain its limit. In the case of the sequence of regular inscribed polygons used in the theorems on the circle, the variable apothem, perimeter, and area do not at- tain their limits. On the other hand variable segments between the sides of a triangle may decrease in length till at the vertex they reach their limit zero ; chords of a circle may increase in length until they reach their limit the diameter. If a point move along a sect in such a manner as to traverse one half of it in a given period of time, one-half the remaining sect in a second period, and so on indef- initely, it will never traverse the entire sect. It is evi- dent, however that by continuing the process for a suf- ficient number of periods of time the portion still un- traversed by the point will become and remain less than any given sect, however small. The entire sect is then the limit of the variable distance traversed by the point. By increasing the number of sides of an inscribed reg- ular polygon ( 408) indefinitely the perimeter of the polygon approaches the circle as its limit and the area of the polygon approaches the area of the circle as its limit, for it is evident that the difference between the 216 PLANE GEOMETRY perimeter and the circle and the difference between the two areas may, by this process, be made less than any as- signed amount. PEOPOSITION I. 426. THEOREM OF LIMITS. // two variables are al- ways equal and each lias a, limit, their limits are equal. i Proof. Let the two equal variables be x and y and let their respective limits be the constants a and b. Rep- resent a x by v and b y by u. Then by subtrac- tion follows (a 2c) (6 $f-) = v u. Since x y this becomes a b = v u. By the definition of a limit ( 405) a and b are constants and therefore a b is a constant. Hence v 11, which equals a & is constant. But since v and u may both be considered as small as one desires and are variables the expression v u can- not be constant unless it is zero in which case a b is zero and a = b. 427. The following statements are here assumed with- out proof, although demonstrations of them are easily made from the definitions on limits. 1. The product of a variable by a constant is a varia- ble and its limit is the product of the constant and MIC- limit of the variable. If the limit of x is a, then the limit of kx is ka, k being a finite C0n8tant ' 2. The quotient of a variable by a constant is a varia- ble and its limit is the quotient of the limit of the varia- ble by the constant. If the limit of x is a, then the limit of ?j- IS T ' fe bein g a K K finite constant different from zero. INCOMMENSURABLE MAGNITUDES 217 PKOPOSITION II. 428. THEOREM. // a line is parallel to the base of a triangle, the sides are divided into propor- tional segments. A The following demonstration is for the case omitted in 251. Given A ACE with BD \\ CE and AB incommensur- able with EC. AB AD To Prove BC DE Proof. SUG. 1. With some unit commensurable with BC divide AB and BC. A remainder, as HB, will occur in the division of AB. If parallel lines are drawn through the points of division DE will be divided into equal segments and the line through H must fall somewhere between the intersection of the preceding parallel line and D, as at I, otherwise HI would not be parallel to the other parallel lines. 2. Then by 251, -- = --. BC DE 3. A sequence of diminishing units makes AH a variable with AB as its limit. Hence 218 PLANE GEOMETRY by 427 - : is a variable with - - as its limit. BC BC Similarly AI is a variable with AD as its limit , AI AD and. - - is a variable with - - as its limit. DE DE 4. By 426 the limits of the two varia- AH AI AB AD bles and are equal and hence BC DE BC DE Therefore PKOPOSITION III. 429. THEOKEM. In the same or in equal cir- cles, angles at the center have the same ratio as their intercepted arcs. The following demonstration is for the case omitted in 297. Given two circles and 0' with Z AOB incommen- surable with respect to Z A'O'B', AB and A'B' being the respective intercepted arcs. ZO arc AB To Prove -r^ = 77^. ZO arc A B Proof. SUG. 1. With an angle commensurable with Z as a unit angle measure the two angles. In Z there will be a remainder as Z MOB which is less than the unit angle, having the intercepted arc MB (the point M always falling between the end of the last unit arc and point B ) . 2 . By , 297 UOM =A*L. LA'O'B' A'B' INCOMMENSURABLE MAGNITUDES 219 3. A sequence of diminishing units makes Z AOM a variable with Z AOB as its limit. Hence by 427 is a variable with as its limit. Similarly arc AM is a va- LA'O'B' riable with arc AB as its limit and - is a va- A'B' 4 B riable with - as its limit. A'B' 4. By 426 the limits of these variable quotients are equal and hence ZO _ Z AOB arc AB ZO'~ZA'0''~arc A'B'' 430. A second demonstration of 429. Assume that Z AOB AB Z AON AB ^ < - and take A so that - (1) LA.'0'B' A'B' LA'O'B' A'B' Measure the given angles by a unit angle less than Z NOB. One point of division will fall then between N and B, say at M and by 297 y^f; = -7^7 (2) LA-(JB A. n From (1) and (2) follows - ='* which is not Z AON AB true since Z AOM > Z AON and arc AM < arc AB. Hence the assumption is false. Similarly it can be shown LAQB . AB that - ' is not greater than --- LA'O'B AB' Therefore Apply this method of proof to Prop. II. 1. What part of the diameter of a circle is the apothem of an inscribed regular triangle? How can this conclusion be used as a basis for inscribing a regular triangle? 220 PLANE GEOMETRY ( PBOPOSITION IV. 4-31. THEOREM. Two rectangles having equal bases are proportional to their altitudes. The following demonstration is for the case omitted in 357. Given two rectangles P and P' with equal bases and altitudes a and a' in- commensurable with respect to each other. To Prove P a Proof. SUG. 1. Measure altitudes a and -a' with a unit commensurable with a'. In the measure- ment of a there will then remain a segment c less than the unit. Why? Through the points of division pass lines parallel to the bases. 2. By 357 Beet. = . P' Alt. a' 3. A sequence of diminishing units makes rectangle AM a variable with the limit P. Hence by 427 the ratio of rectangle AM to P' p is a variable with as its limit. Similarly al- titude AM is a variable with a as its limit and hence the ratio of altitude AM to a' is a variable with as its limit. a' INCOMMENSURABLE MAGNITUDES 221 4. By 426 the limits of these varia- bles are equal and hence = P' a' Therefore^ Adapt the demonstration of 430 to this theorem. PKOPOSITION V. 432. THEOREM. Two circles are to each other as their radii. This proof of 409 is based on the theory of limits. Given two circles c and c' with radii r and r' respect- ively. To Prove - = - c' r' Proof. SUG. 1. Inscribe in the circles c andV sim- ilar regular polygons with respective perimeters v r T p and p' . Then = and p = p' x p r' r' 2. By increasing the number of sides p becomes a variable with c as its limit and p' be- comes a variable with c' as its limit. Hence by 407 the product p' x i s a variable with r' Cf X; I as its limit. r f 3. By 426 the limits of these varia- bles are equal and -c = c' X and hence = r' c' r' Therefore 1. A regular hexagon with 6 in. side is inscribed in a circle. Find the area of the regular inscribed triangle. 2, If the angle of a parallelogram is bisected and the bisector extended to the opposite side an isosceles triangle is formed. 222 PLANE GEOMETRY PROPOSITION VI. 433. THEOREM. The areas of two circles are to each other as the squares of their radii. (412) Prove this theorem in a manner like that of 432. 434. A second proof of 432. SUG. 1. Inscribe in c and c' similar regular polygons with perimeters TJ T p and p' respectively. Then = Why? p' r C )* )' 2. Assume > and hence c > c' X . c' r' r' Also p < c and p' < c'. 396. 3. Since in the sequence of inscribed poly- gons there is a point at which the inscribed poly- gon is nearer to c than any assigned number, consider a polygon q inscribed in c so that T q> c' X - Let q' be the similar polygon in- r' scribed in c'. Then = > and q q' X . q' r' r 4. Since q' < c' it follows that Y T T q' x < c' X and hence q < c' X r r' r' But as this is contrary to fact the assumption in step 2 is false. 5. Similarly prove that cannot be c T less than r' Therefore 435. Adapt, the demonstration 434 to theorem 433. SOLID GEOMETRY. CHAPTER VII LINES AND PLANES 436. SOLID GEOMETRY. That portion of Geometry which treats of figures the parts of which are not con- fined to a single plane (19) is Solid Geometry. The resultc of plane geometry furnish the basis for investiga- tions in solid geometry, but it is to be remembered that the state- ments of plane geometry are made with respect to figures which are entirely in one plane and are not necessarily true in solid geometry. For instance, it has been proved in plane geometry that only one perpendicular can be erected to a line at a given point, but in solid geometry many perpendiculars can be so erected. This may be illustrated by the spokes of a wheel all of which are perpendicular to its axis. Therefore in solid geometry the theorems of plane geometry must not be applied unless the reference is to parts of a figure all of which are in one plane; but such theorems may be applied first to one plane, then to another, and so on. The pupil will be much helped in the study of solid geometry by notic- ing that most of the theorems are but extensions or generalizations of theorems previously studied in the plane geometry. 437. The relations of the parts of a figure or figures in a plane are not changed by moving the plane con- taining them from one position to another. 50 (4). 438. A PLANE. A surface such that the straight line joining any two of its points lies entirely in the surface is a plane. (18.) A plane is unlimited in extent and from the definition it follows that if a straight line of a plane be indefinitely extended, it can never leave the 224 SOLID GEOMETRY t plane. A plane embraces a line, or is passed through a line, when the line lies wholly in the plane. 439. INTERSECTION OF PLANES. That portion of two planes which is common to both is their intersection. 440. PLANE DETERMINED. A plane is determined by certain lines or points when no other plane can embrace those lines or points without coinciding with the first plane. 441. POSTULATE. A plane can be revolved about a line as an axis. Hence it may be inferred that a plane while embrac- ing a line can take an infinite number of positions; and that, as a plane is unlimited in extent, all points in space can be embraced by the plane in the course of one com- plete revolution. PROPOSITION I. 442. THEOREM. A plane is determined I. By a straight line and a point without the line. II. By three points not in a straight line. III. By two intersecting straight lines. IV. By two parallel straight lines. I. Given, the straight line EF and the point P not in EF. To Prove that EF and P determine a plane. Proof. SUG. 1. Through line EF pass a plane and revolve it about EF as an axis until it contains LINES & PLANES 225 the point P. 441. 2. How much, can the plane be re- volved either way about EF and still contain point P? Why? 4,5. 3. How many planes can embrace the given point and given line? 4. ' EF and P determine the plane MN. 440. Therefore II. Given points D, E, F not in a straight line. To Prove that D, E, F determine a plane. Proof. SUG. Connect two of the points, and com- plete the demonstration. Therefore- Ill. Given DE and FG, two intersecting lines. To Prove that DE and FG determine a plane. Proof. SUG. 1. Since the lines may be unlimited in extent, let E be any particular point in DE other than 0. Then FG and E determine a plane. Why? 2. This plane contains the given lines. Why? 3. There is but one such, for if two planes contained the given lines they would each 226 SOLID GEOMETRY contain FG and E. This is impossible. Case I. Therefore N- IV. Given DE and FG, two parallel lines. To Prove that DE and FG determine a plane. Proof. SUG. 1. By definition of parallel lines there is at least one plane containing these lines. Rep- resent it by N. 2. If there is more than one plane through these lines, there will be more than one plane containing DE and F, a point on FG. This is impossible by Case I. Therefore 1. A straight line can intersect a plane in but one point. Sue. Suppose two points of the line to be on the plane. 2. Three straight lines each intersecting the two others, but not in a common point, lie in a plane. 3. What is the greatest number of planes that may be deter- mined by two intersecting lines and a point? By three parallel lines? 4. A carpenter wishing to determine whether a board or other surface is a plane, places a straight edge (try-square) upon it in various positions. What is the test of the straight edge and why does it determine the question involved? 5. Can two lines be so placed as not to lie in one plane? 6. Can four points be so placed as not to lie in one plane? Give half a dozen illustrations of mechanical difficulties growing out of the answer. 7. Which is the more likely to stand firm, a three or a four legged stool? Why? 443. POSTULATE. Two interceding planes have at least two points in common. LINES & PLANES 227 PROPOSITION II. 444. THEOREM. The intersection of two plane* is a straight line. V Given two intersecting planes, M and X with two points E and G in common. 443. To Prove that the intersection of M and N is the straight line EG. 439. Proof. SUG. 1. Where does line EG lie with respect to each plane? Why? 2. Let be any point in plane M out- side of line EG. Can lie in plane Nf Why? 442. 3. What are the only points common to planes M and X .' Therefore 1. What is the locus of a point common to two planes? 445. THE FOOT OF A LINE. The point in which a line meets a plane is the foot of the line. 446. PERPENDICULAR TO A PLANE. A line is perpen- dicular to a plane when it is perpendicular to every line in the plane passing through its foot. The plane is then said to be perpendicular to the line. 447. OBLIQUE TO A PLANE. When a line is oblique to one or more lines of a plane, it is oblique to the plane. 2. Why is the crease formed in folding a piece of paper for an envelope a straight line? 228 SOLID GEOMETRY PROPOSITION III. 448. THEOREM. // a straight line is perpen- dicular to two lines of a plane at their point of intersection, it is perpendicular to the plane. Given EO -L AB and CD at 0, and plane M determined by AB and CD. To Prove EO _L plane M. Proof. SUG. 1. Extend OE to P, making OP=OE, and let OH be any line of the plane through 0. Draw BD and let H be the point in which it meets OH. Connect both E and P with the points B, H, and D. 2. In the A BEP compare BE and BP. In A DEP compare DE and DP f 73. 3. Compare A EBD and PBD; .A. EBD and PBD. Auth. 4. Compare A EBH and P## ; lines EH and Pfl". Auth. 5. Compare A #0# and POfi"; 1 #0# and FOfl". What relation does EO bear to OH ? Or relate #0 to OH by 76. Therefore LINES & PLANES 229 449. COR. At a point in a plane only one perpendic- ular to the plane can fee erected. Given EF 1 plane M at point F. To Prove EF the only perpendicular to M at F. Proof. If another perpendicular can be erected, rep- resent it by DF. The lines EF and DF determine a plane ( 442) which will intersect plane M in a straight line as PF. Then in this plane, both EF and DF are per- pendicular to PF at point F. Is this possible ? Why ? Therefore PEOPOSITION IV. 450. THEOREM. From a point to a plane there is one line which is shorter than any other. Given, a plane M and a point P without the plane. To Prove that from P to plane M there is one line shorter than any other. Proof. SUG. 1. If there is not one shortest line, there must be a group of equal shortest lines. Let PE and PF represent two such lines. Con- 230 SOLID GEOMETRY nect E and F and let be the mid-point of EF: Join P and 0. 2. With respect to length, compare PO with PE or PF. Therefore 451. COR. I. The shortest line from a point to a plane is the perpendicular from that point to the plane. Given PO the shortest line from point P to the plane M. To Prove that PO is per- pendicular to plane M. Proof. Through the foot of PO draw in plane M any two lines as EF and GH. Then PO 1 EF and PO 1 GH. Why? /. PO 1 plane M. Why? Therefore 452. COR. II. From a point without a plane only one perpen- dicular can be dropped to the plane. Given PO 1 plane N from point P without the plane. Proof. If there is a second perpendicular from P to N, represent it by PG. Then the plane PGO ( 442) intersects plane N in line OG. Why? What relation does PG bear to line OG1 Why? What relation does PG bear to plane N ? Why ? Therefore 453. DISTANCE FROM A POINT TO A PLANE. The length of the perpendicular from a point to a plane is the distance from the point to the plane. LINES & PLANES PBOPOSITION V. 454. THEOREM. All perpendiculars to a line at a given point lie in one plane perpendicular to tlie line at that point. Given plane M 1 PO at and line OH any line l PO at 0. To Prove that OH lies in plane M. Proof. SUG. 1. The plane N determined by PO and OH will intersect plane M in a straight line through 0, as OG. Why? 2. POA.OG. Why? 3. OG and OH both lie in plane N and must therefore coincide. Why? 4. ' OH must lie in plane M. Why? 5. As Ofl" was any line perpendicular to PO at 0, the same is true of all such lines. Therefore 455. COB. I. Through a point in a line only plane can be erected perpendicular to the line. Given a line a and a plane M 1 to a at point 0. To Prove that M is the only plane 1 a through O. Proof. Suppose that there is a second plane, as N, -L a at 0, and let b be the intersection of planes M and N. Pass a plane P through line a which does not contain line b. This plane will cut M and A 7 in one 232 SOLID GEOMETRY two distinct lines, each perpendicular to a. But this, by 454, is impossible. Why? 456. COR. II. From a point without a line only one plane can be drawn perpendicular to the line. Given point P not on line a and plane M through P and 1 a. To Prove that M is the only plane through P and 1 a. Proof. Let M meet line a in point A. If there is a, second such plane, as N, let it meet a in point B. Pass a plane through P and line AB or a. This plane will cut M and N, each in a straight line. Complete the demonstration. 1. Erect a plane perpendicular to a line at a given point. 454. 2. How can a carpenter erect a studding perpendicular to a floor with his square? 3. Set up in the school room a pole perpendicular to the floor, using a right angle (a carpenter's square or a book cover). 448. 4. How many braces are required to hold the pole of the preceding exercise in position? 5. Construct a plane perpendicular to a given line from a point without the line. 448. 6. If a plane is perpendicular to a sect at its mid-point every point in the plane is equidistant from the extremities of the sect and every point outside the plane is not equidistant from the extrem- ities. LINES & PLANES 233 Sue. Let X be any point in plane M, which is perpen- dicular to sect DE at its mid point 0. Prove DX = EX. PROPOSITION VI. 457. PROBLEM. To drop a perpendicular from a point to a plane. Given a plane M and point P not on M. To Construct a line PO from PA.M. Construction. SUG. 1. Draw any line EF in plane M and drop a perpendicular, PH, to EF in the plane of EF and P. Auth. 2. Through H in plane M draw HG1.EF. Auth. 3. From P drop POA.HG. Auth. 4. To prove PO 1 M, draw a line from to 1, any point in EF except H, and join P and J. 5. ~PH 2 -OH 2 =P0 2 . Auth. 6. EH 2 +On 2 =OE 2 . Auth. 7. Adding, #P 8 = P0 8 +0# 2 - S. ' POOE. 9. * POlM. Why? 1 . What is the locus of points in space equidistant from two given points? 2;U SOLID GEOMETRY ( 458. Locus IN SPACE. The locus of a point in space satisfying certain given conditions consist of those geo- metric figures in space to which the point is limited and every point of which satisfies the conditions. 164-165. PROPOSITION VII. 459. PROBLEM. To erect a perpendicular to a plane at a given point in the plane. Given plane M and point in M. To Construct OP LM at 0. Construction. Sue. 1. In plane M draw a line a and then two lines c and d each perpendicular to a at 0. 2. What relation does the plane of c and d bear to a ? 3. This plane cuts M in line I). Why? 4. In this new plane draw OP 1 1). Auth. How is OP related to af Why ? 5. How is OP related to plane M? Why? Compare this construction with ex. 3, p. 232. 1. Another demonstration for 448. Draw from D a line DB terminated in line AB and bisected by OH at H. Auth. (Use the fig. of 44.8.)_ EB Z +EI) 2 = 2EH* + 2 #T a QB z +f)J)* = 20H*+ 2 BE 2 345 Subtract and reduce. Apply 311. LINES & PLANES 235 PROPOSITION VIII. 460. THEOREM. If , from the foot of a perpen- dicular to a plane, a line is drawr, perpendicular to any given line in the plane and, from this point of intersection, a line is drawn to any point of the perpendicular to the plane, the last line is perpendicular to the given line in the plane. Given line EG 1 plane M and CD any line in M, with EO 1 CD and F any point in EG. To Prove OF 1 CD. Proof. SUG. 1. On line CD take OD = OC. Join F with C, D and 0, and E with C and D. 2. In &ECD compare EC and ED. Autli. 3. Compare FC and FD. 65. 4. What relation does OF bear 76. to CD? Therefore 1. In the fig. of 460, given EG L EO, EO LCD, and OF LCD at 0, F being in EG. Prove EG 1 plane M of EO and CZ>. SUG. EF* = OF^ 02 and EC* = 0(72 + 0#2. Add and complete the demonstration. 2. Show that by marking at right angles to the edges of a stick of timber with a carpenter's square continuously the line will end at its starting point. 236 SOLID GEOMETRY PROPOSITION IX. 461. THEOREM. // one of two parallel lines is perpendicular to a plane, the other is also perpen- dicular to the plane. H D Given GH II CD and meeting plane M in G and C re- spectively, with CD j_ M. To Prove GHi_M. Proof. SUG. 1. GH and CD determine a plane which meets plane M in GC. Why ? Join G witk O y any point in CD. Draw EF in plane 3f and 2. What relation does EF sustain to GO? (460.) ToGCf To plane #C7 448. 3. What relation does EF sustain to GH, or GH to #jF? # to GC f 90. H to plane M? Why? PROPOSITION X. 462. THEOREM. Two Zmes perpendicular to the same plane are parallel. Given two lines a and &, each perpendicular to plane M. To Prove a II b. LINES & PLANES 237 Proof. SUG. 1. Suppose a is not parallel to b and through a point on a draw a line c I! 6. 2. What relation does c bear to plane Mt Why? 461. 3. Complete the demonstration. Therefore PROPOSITION XL 463. THEOREM. Two straight lines each par- allel to a third straight line are parallel to each other. 7 Given a II c and b II c. To Prove a II 6. Proof. SUG. 1. Construct plane M .c. Auth. 2. What relation do a and b bear to Mf Auth. 3. What relation do a and b bear to each other? Auth. Therefore ]. To draw a straight line which shall intersect each of two lines not in the same plane and shall pass through a point not in either line. SUG. Pass a plane through the given point P and one line a. Also pass a plane 9 through P and the second line ft. These ,s planes intersect in a line x through P. Why? This line x also meets both a and &. Why? 464. PROJECTION OF A POINT ON A PLANE. The foot of the perpendicular from a point to a plane is the pro- jection of the point on the plane. 238 SOLID GEOMETRY e 465. PROJECTION OF A LINE UPON A PLANE. The locus of the projections of the points of a line upon a plane is the projection* of the line upon the plane. PROPOSITION XII. 466. THEOREM. The projection of a straight line upon a plane is a straight line. Given a plane M and a straight line EF not in M. with E' and F' the projections of E and F respectively upon M. To Prove the straight line E' F' the projection of EF. Proof. SUG. 1. It is necessary and sufficient to show that the projection P' of any third point in EF lies in E'F'. 2. Since PP' , EE' , FF' are each l_M, they are parallel. Auth. 3. EE' and FF' determine a plane, A T , cutting M in line E'F'. Why? 4. In plane N, it is possible to draw a line through P which is parallel to FF' and hence 1 M, but this line must coincide with PP', Why? 5. Since PP' lies in N, its foot in M must lie in E'F' , Therefore 1. The area of a circle is 20 sq. in. What is the area of a circle of double the radius? Of ^ the radius? Of m times the radius? Of times the radius? m 2. If a is the radius of a circle with an area of 40 sq. in., what is the radius of a circle of 80 sq. in.? Of 10 sq. in.? Of ICO sq. in.? LINES & PLANES 239 PROPOSITION XIII. 467. THEOREM. // from a point to a plane the perpendicular and oblique lines be drawn I. The perpendicular is the shortest line from the point to the plane; II. Oblique lines having equal projections upon the plane are equal; III. // two oblique lines are equal, their pro- jections upon the plane are equal; IV. Of two unequal oblique lines, the line hav- ing the greater projection is the greater; V. The greater of two oblique lines has the greater projection upon the plane. I. Given PO 1 plane M. To Prove PO shorter than any other line from P to .V. Proof. SUG. Compare A POG and PGO. II. Given PG and PE oblique to M with their re- spective projections OG and OE equal. To Prove PG PE. Proof. The demonstration is left to the pupil. III. Given PG = PE, with OG and OE their re?pec- tive projections. To Prove OG = OE. Proof. The demonstration is left to the pupil. 240 SOLID GEOMETRY IV. Given obliques lines PE and PF, with OE and OF their repective projections, and OE > OF. To Prove PE > PF. Proof. SUG. 1. On OE take 021 = 0^ and draw PH. 2. Compare PE with P#; P# with PF. Complete the demonstration. V. Given PE > PF. To Prove OE > OF. Proof. SUG. Use the indirect method. Therefore 1, What is the locus of the foot of an oblique line in a plane when the oblique line is revolved with one end not in the plane stationary? 2, An oblique to a plane intersects its own projection upon the plane. 3, The locus of a point in space equidistant from all points in a circle is a straight line through the center of the circle and perpendicular to its plane. 4, A ceiling is 8 ft. high. A pole 10 ft. long is held at a given point on the ceiling and revolved with the lower end touch- ing the floor. Find the diameter of the circle described. 5, How could the pole in the preceding exercise be used to find that point on the floor which is directly under the point on the ceiling? 6, Show that the angle that the pole makes with the perpen- dicular is constant as the pole revolves. 7, A horse is tied with a rope 100 ft. long to a point on a vertical pole 25 ft. from the ground. What is the shape of his pasture and how far can he get from the foot of the pole? 468. LINE AND PLANE PARALLEL. A straight line and a plane are parallel if they can never meet, however far they may be extended. 469. PARALLEL PLANES. Two planes which cannot meet, however far they may be extended, are parallel planes. LINES & PLANES 241 PROPOSITION XIV. 470. THEOREM. A straight line is parallel to a plane if it is parallel to a line in the plane. Given line a outside plane M and parallel to line b in plane M . To Prove a II M. Proof. SUG. 1. a and b determine a plane N. Why? 442. 2. What is the intersection of M and Nf Why? 444. 3. If a and M have a point in common it must lie in b. Why? 444. 4. This is impossible. Why ? 438. Therefore 471. COR. I. A plane may be passed through one of two non-intersecting lines parallel to the other. Given two non-intersecting lines a and b. To Prove that a plane, M, can be passed through b parallel to a. Proof. SUG. From any point in b draw the line c parallel to a. Then plane M determined by b and c is parallel to a. Why? 1. How many such planes (471) are there? 2. Jf a line is parallel to a plane, it is parallel to its own projection upon the plane. 242 SOLID GEOMETRY 472. COR. II. Through a point not on either of two lines a plane can bo passed parallel to both the lines. A X~~- p. x' Given lines a and b and point P. To Prove that a plane M parallel to a and to b can be passed through P. Proof. Through P draw two lines, c and d, parallel to a and b respectively. Complete the demonstration. PEOPOSITION XV. 473. THEOEEM. // a line is parallel to a plane f any plane that embraces the line and intersects the plane intersects it in a line parallel to the given line. N < Given line c parallel to plane M, and plane N through c and intersecting M in line b. To Prove b I! c. Proof. SUG. 1. If c and b are not parallel they will meet at some point as 0. Why? This is impos- sible. Why? Therefore 1. Through a given point in space one and only one line can be drawn parallel to a given line. 87. 2. Through a given point in space any number of lines can be drawn parallel to a given plane. 3. All lines through a given point and parallel to a given plane lie in one plane. SUG. Drop a 1 from the given point to the given plane. 4. Through a given point only one plane can be passed parallel to a given plane. LINES & PLANES 243 PROPOSITION XVI. 474. THEOREM. Planes perpendicular to the same straight line are par alb I. \ Given two planes M and A" each f~~ ~~} 1 perpendicular to line a. / _J / To Prove M I! A 7 . Proof. SUG. Use an indirect proof /~~ based on 450. /_ Therefore PROPOSITION XVII. 475. THEOREM. // two intersecting lines in one plane are each parallel to a second plane, the planes are parallel. 7 7 Given two lines a and 6 in plane M, meeting in O, each line parallel to plane A T . To Prove M II N. Proof. SUG. 1. At 0, the common point of a and 1), erect a 1 to plane M and extend it to meet plane N at 0'. Line 00' is l to a and to 6. Why ? 2. Lines a and 00' determine a plane which intersects N in a line as a' through 0' . Likewise b and 00' determine a plane which in- tersects A 7 in &'. Auth. 3. What relation does a' bear to ? ?/ to 1} ? Auth. 244 SOLID GEOMETRY 4. What relation does 00' bear to a' and to &'? Auth. 5. What relation does 00' bear to plane N? Auth. 6. What relation does N bear to Mf Auth. Therefore PROPOSITION XVIII. 476. A line perpendicular to one of two par- allel planes is perpendicular to the other. Given plane M II plane N and line a 1 M at 0. To Prove a_LN. Proof. SUG. 1. Through any point in plane M draw two lines as b and c. Through a and & and through a and c pass planes and let these two planes intersect plane N in lines b' and c' re- spectively. 2. What relation does &' bear to b? c' to c? Auth. 3. What relation does a bear to b and to c? To b' and to c' ? Auth. 4. What relation does a hoar to Nf Auth. Therefore 1. Two planes parallel to the same plane are parallel to each other. 474, 476. 2. What is the locus of the foot of an oblique line 10 in. long drawn to a plane from a point eight in. from the plane? Compute the area of the figure bounded by the locus. LINES & PLANES 245 477. DISTANCE BETWEEN PARALLEL PLANES. The length of the sect intercepted between two parallel planes and perpendicular to both is the distance be- tween the parallel planes. PROPOSITION XIX. 478. THEOREM. // two parallel planes are cut by a third plane the intersections are parallel. Given two parallel planes M and N, cut by a third plane P in lines m and n respectively. To Prove m II n. Proof. SUG. Show that m and n lie in the same plane and cannot meet. Therefore 479. COR. I. Parallel lines intercepted between parallel planes are equal. SUG. Assume lines a and & in 478 to be parallel. Prove them equal. 480. COR. II. Parallel planes are everywhere equi- distant. 1. A straight line and a plane both perpendicular to the same straight line are parallel. 2. What is the locus of a line through a given point and parallel to a given plane? 3. If a straight line and a plane are parallel, any line par- allel to the given line is parallel to the plane also. 246 SOLID GEOMETRY t PROPOSITION XX. 481. THEOREM. // two angles, not in the same plane, have their respective sides parallel and ex- tending in the same directions from the vertices, the angles are equal. Given Z BAG in plane M and Z B'A'C' in plane N, with A#IIA'l?' and ACllA'C', the respective direc- tions from A and A ' being the same. To Prove Z BAC= Z 'A'C'. Proof. SUG. 1. Take points B, B', C, C' so that AB = A'B' and AC = A'C'. Connect A and A', B and 5', C and C', 5 and C, B' and C". 2. Compare AA' and BB\ A A' and CC', BB' and CC". 3. Compare BC and 'C"; ABAC and A B'A'C 1 ; Z A and Z A'. Auth. Therefore 1. What is the locus of a point at a given distance from a given plane? 2. What is the locus of a point equidistant from two given points and also at a given distance from a given plane? When is there no solution? 3. From A, the mid-point of one side of a parallelogram draw lines dividing each of the adjacent sides into three equal parts and the opposite side into six equal parts. Prove that each of the twelve triangles thus formed equals one twelfth of the parallelogram. LINES & PLANES 247 PROPOSITION XXI. 482. THEOREM. // three parallel planes inter- sect two straight lines, the corresponding seg- ments are proportional. Given three parallel planes M, N, P intersecting the two lines a and a' in the points D, E, F and D', E' , F' respectively. _ _ DE D'E' To Prove = EF E'F' Proof. SUG. 1. Join D and F' and let G be the in- tersection of this line with plane N. 2. Plane DD'F' intersects planes M and N in what lines? Plane DFF' intersects planes A r and P in what lines ? 3. What relation does EG bear to FF' ? E'GtoDD'1 Why? . DE DG D'E' 4. Compare the ratios 251. 5. Complete the demonstration. Therefore 1, If two parallel planes intersect two parallel planes the four lines of intersection are parallel. 2. What is the locus of a point in space equidistant from two parallel planes? 248 SOLID GEOMETRY 1. Find a point X equidistant from two given points, equi- distant from two other given points, and in a given plane. Is the problem ever impossible? Will any arrangement of the given parts permit an unlimited number of solutions'? 2. Another demonstration for 461. Given = AB \\ A'B' and A'B'l. plane M. To PROVE AB 1 plane M. SUG. Draw in plane M any two lines through B' as B'C' andB'D'. Through B draw BC and BD parallel to B'C' and B'D' respectively. &. A . A' B'C' and A' B'D' are right angles. Why? They are equal to ^ AB and ABD respectively. Why! Therefore ABLM. Why? 3. Find a point X equidistant from two given points, equi- distant from two given planes, and at a given distance from a third plane. Discuss all possibilities. 4. What is the locus of a point equidistant from two given points A arid B and also from two given points C and D1 Discuss all possibilities due to the different positions of the pairs of points. 483. DIHEDRAL ANGLE. r iwo planes which meet or intersect form a dihedral angle. The two planes M and N meeting in the line ED form a dihedral angle. The intersecting planes, M and A 7 , are the faces of the angle and the line of intersection, ED, is the edge. A dihedral angle is read by reading in order one face, the edge, and the second face. When but one dihedral angle is formed at an edge, the angle may be read by naming the edge, as dihedral angle ED. A dihedral angle is often called a dihedral. 484. PLANE ANGLE OF A DIHEDRAL. An angle formed by two lines, one in each face of a dihedral, perpendicu- lar to the edge at the same point is the plane angle of the dihedral Lines FO and HO lying in the faces M DIHEDRAL ANGLES 249 and A T respectively and each perpendicular to KI) at O is a plane angle of the dihedral ED. 485. COR. I. The plane angle of a dihe- dral is the same size from whatever point of the edge it is drawn. Given dihedral EE' with plane angles E > GEH andG'E'H'. To Prove L GEH= L G'E'H'. Proof. Left to the pupil. 481. 486. COR. II. The angle formed by the intersections of the faces of a dihedral with a plane perpendicular to the edge is a plane angle of the dihedral. Proof left to the pupil. 484. 487. EQUAL DIHEDRALS. Two dihedral angles are equal when they can be made to coincide. The magnitude of a dihedral does not depend upon the extent of its faces. If a plane be made to revolve about the edge as an axis from the position of one face to the position of the other face, it revolves or turns through the dihedral angle and the greater the amount of the turning the greater the angle. 1. A and B are two points equally distant from a plane and upon the same side of it; prove that line AB is parallel to the plane. 2. Determine a point E in a plane such that the difference between its distances from two given points on opposite sides of the plane is a maximum. Sue. Drop a perpendicular to the plane from one of the points as H and extend it to G, an equal distance on the other side of the plane. Connect G with the second point F, this line meeting the plane in E. Prove E to be the required point, as follows : EU EF = FG. Take any other point as E' in the given plane, join it to F and H and prove FG > E'H E'F. 250 SOLID GEOMETRY 1. Determine a point E in a plane such that the sum of its distances from two points on the same side of the plane is the minimum. Sue. Drop a perpendicular to the plane from one of the points as H and extend it to G, an equal distance beyond the plane. Join H G to the second point F, this line j^s^; -._/% meeting the plane in E. Prove E to be the required point. To do this, take any other point as E' in the //-" given plane and prove HE' + E'F c > HE + EF. PEOPOSITION XXII. 488. THEOREM. Two dihedral angles are equal if their plane angles are equal. H' Given two dihedrals GH and . take BD BC and join A and D. 3. Compare AC with AD. Auth. 4. Compare L ABC with LAUD. 124. Therefore ' DIHEDEAL ANGLES 259 1. The locus of points in space equidistant from two inter- secting planes is the pair of planes bisecting the dihedrals formed by the given planes. 503. ANGLE OF A LINE TO A PLANE. The acute angle which a line makes with its projection on a plane is the angle of the line to the plane. 2. The supplement of the angle of a line to a plane is the greatest angle which the line makes with any line in the plane through its foot. 3. A line makes equal angles with parallel planes. 4. There is one and but one perpendicular which can be drawn to two non parallel lines which are not in the same plane. Given two lines, b and c, not in the same plane and not parallel. Pass a line d through one of them, as b, parallel to the other, c. Pass through b and d the plane M. How does M lie with reference to line c? Through c pass a plane N per- pendicular to M. Let line e be the inter- section of M and N. Then e \\ c. Why? Let the intersection of b and e be and at erect the line OPLM. This line lies in N. Why? Hence OP meets c and OP 1 c. Why? THEREFORE there is at least one such common perpendicular to b and c. Suppose there is another such perpendicular as EF. The plane of line c and EF will intersect If in a line GF and be 1 to M. Why? Thus there will be two planes embracing c and each i M, -\\uich is impossible? Why? THEREFORE 5. In the preceding theorem the common perpendicular, OP, is the shortest line which can be drawn between the two given lines. 6. Find a point X equidistant from four given points not in the same plane. Sue. Let the given points be A, B, C, D. What is the locus of points equidistant from A and B ? From B and C? 260 SOLID GEOMETRY I From A, B, and C ? From C and D Complete the demon- stration. 504. POLYHEDRAL ANGLE. Three or more planes meeting at a common point form a polyhedral angle; or simply a polyhedral. For example, A B(JD represents a polyhedral angle formed by three planes. The common point, A, is the vertex of the angle, the intersections of the planes, AB, AC, AD, are the edges of the angle, and the portions of the planes included between the successive edo-es* are the faces of the angle. The plane angles formed by the successive edges are the face angles of the polyhedral. The face angles and the dihedrals of the successive faces are the parts of the polyhedral. 505. CONVEX POLYHEDRAL ANGLE. If the intersec- tions of a plane with all the faces of a polyhedral form a convex polygon the polyhedral is a convex polyhedral angle. 506. CLASSIFICATION OF POLYHEDRAL ANGLES. A poly- hedral with three faces is a trihedral angle, or a trihe- dral; one having four faces is a tetrahedral angle, or tetrahedral; etc. A trihedral angle with two equal face angles is isosceles. 507. CONGRUENT POLYHEDRAL ANGLES. Polyhedral angles which can be made to coincide are congruent poly- hedral angles. For two polyhedrals to be congruent, it is necessary that the respective parts of the two angles be equal and arranged in the same order. 508. SYMMETRICAL POLYHEDRAL ANGLES. Two poly- hedral angles having the dihedrals ai^d faces angles of the one equal respectively to the corresponding parts of the other but arranged in the reverse order are symme- trical polyhedral angles. POLYHEDRAL ANGLES E' 261 In general two symmetrical polyhedral angles cannot be made to coincide. Polyhedrals E FGH and E' F'G'H' are symmetrical pro- vided L PEG = L F'E'G', L GEE - L G'E'H', L HEF = L H'E'F', dihedral EF dihedral E'F', etc., the arrangements of the re- spective parts in the two polyhedrals being opposite. As an illustration of the symmetrical relation consider a pair of gloves. Two gloves for the same hand may be compared to equal polyhedrals and the pair to symmetrical polyhedrals. 1. Cut from pasteboard two figures like those in the ac- companying figures, creasing them on the dotted lines. Note that all parts with corresponding notations are equal. The equal- ity of the "parts" may be tested by superposition. The figures when bent into polyhedrals will not coincide. E 2. If a plane be passed through either diagonal of a par- allelogram, the perpendiculars to this plane from the extremities of the other diagonal are equal. 3. Having given a fixed straight line and two points' not in the line, find a point in the fixed line equally distant from the given points. 4. What is the locus of a point equidistant from two given parallel planes and at the same time equidistant from two given points ? 262 SOLID GEOMETRY ( PROPOSITION XXX. 509. THEOREM. The sum of any two face an- gles of a trihedral angle is greater than the third. Given the trihedral A BCD, in which face angle DAC is the greatest. To Prove Z CAB + Z BAD > L DAC. Proof. SUG. 1. It is unnecessary to prove the theo- rem for the cases in which the greatest angle, LDACy is one of the two angles added. Why? 2. In the face DAC draw a line AM equal to AB f making Z MAD equal to Z DAB, and through B and M pass a plane cutting the two other edges in 7 points D and C. 3. Compare A BAD with A MAD; DM with.DB. Auth. ' 4. Compare DB + BC and BM + MC, compare BC and MC. Auth. 5. Compare Z BAC with Z MAC. Auth. 6. Compare /. BAC + /. BAD with Z DAC. Therefore 1. What is the locus of a point equidistant from two given parallel planes and at the same time equidistant from two other parallel planes? POLYHEDRAL ANGLES 263 PROPOSITION XXXI. 510. THEOREM. The sum of the face angles of any convex polyhedral angle is less than four right angles. Given a polyhedral angle A with n face angles. To Prove the sum of the face angles about vertex A to be less than 4 right angles. Proof. SUG. 1. Pass a plane through the polyhe- dral cutting all the edges in the points B, C, D, F, etc. This cross section polygon will have n sides and the plane will form with the faces n "face" triangles with a common vertex A. 2. Let be any point within this poly- gon and join to the n vertices of the polygon, forming n "base" triangles with a common ver- tex 0. 3. The sum of the angles of the face triangles equals the sum of the angles of the base triangles. Why? 4. Z ABC + L ABF > L CBF, L ACS + L ACD > L BCD, etc. 5. Compare then the sum of all the base angles of the face A with the sum of all the base angles of the base A. 6. Compare the sum of the face angles at A with the sum of the angles about 0. 264 SOLID GEOMETRY 7. Compare the sum of the angles at A with four rt. A. Therefore PROPOSITION XXXII. 511. THEOREM. // two trihedral angles have the three face angles of the one equal respectively to the three face angles of the other, the corre- sponding dihedral angles are equal. Given two trihedrals, A and A', with Z BAG = L B'A'C', L CAD = L C'A'D', L DAB = Z D'A'B'. To Prove dihedral AB = dhl A'B', dhl AC = dhl A'C', and dhl AD = dhlA'D'. Proof. SUG. 1. Take points on the six edges so that AB AC = AD = A'B' = A'C' = A'D' and AM = A'M'. Through B, C, D and B', C', D' pass planes. In the faces BAC and BAD re- spectively draw MN and MP each l AB. MP and MN can meet BC and Z> in N and P respectively for A. AB C and AB Dare acute. Why? Similarly draw lines M'N' and M'P' through point M'. 2. The dihedrals AB and 4' 5' are equal if Z PM#= Z P'M'N'. Why? 3. Compare A AB(7 with AA'B'C',- L ABC with Z A'B'C', BC with B'C'. Auth. POLYHEDRAL ANGLES 265 4. Draw the similar conclusions from &ABD and A'B'D'. Also from A ACD and A'C'D'. 5. Compare A BMP with kB'M'P', BP with B'P' ; MP with J/'P'. Auth. Draw similar conclusions from A BMN and A B'M'N'. 6. Compare A BCD with AB'C'D'; Z C5D with Z C'B'D'; A PA 7 with A tf'P'tf'; AT with N'P'. Auth 's. 7. Compare & XMP with A N'M'P', Z J/ with Z jl/'. Auth. 8. Compare dihedral AB with dihedral A''. Auth. 9. By similar arguments compare di- hedrals AC and A'C", AD and A'ZT. Therefore PROPOSITION XXXIII. 512. THEOREM. // two trihedral angles have the three face angles of the one equal respectively to the three face angles of the other, they are cither congruent or symmetrical. Given two trihedrals, and 0', with ZJfOP Z.'0'P', ZPON = Z.P'0'N', and /.NOM = Z.V'OMT. I. To Prove trihedrals and 0' congruent, the an- gles being arranged in the same order. 266 SOLID GEOMETRY Proof. SUG. 1. Place upon 0' so that /.MOP coincides with LWO'P' and the two edges ON and O'N' are on the same side of the plane MOP. 2. How does plane PON lie with ref- erence to plane P'O'N't Plane NOM with ref- erence to plane N'O'M'f 511. 3. Where does line ON lie with respect to O'N"! Why? II. To Prove trihedrals and 0' symmetrical, the angles being arranged in reverse order. Proof. By 511 the corresponding dihedrals are equal. The parts are therefore respectively equal and by hypothesis the order of the angles is different in and 0'. Therefore 508. If a line and a plane are parallel, a line drawn from any point in the plane parallel to the line will lie in the plane. 1. Parallel lines intersecting a plane make equal angles with the plane. 2. What is the locus in space of points equidistant from the sides of a plane angle? 3. Bisect a dihedral angle. 4. What is the locus of a point in a plane equidistant from two points without the plane? 5. Find a point X in a given line and equidistant from two points without the line. 6. Find a point X in a plane and equidistant from three points without the plane. 7. What is the locus of a point X in a given plane, a given distance from a point not in that plane? 8. Find a point X in a plane and equidistant from the three edges of a trihedral. POLYHEDRAL ANGLES 267 PROPOSITION XIV. 513. THEOREM. Two symmetrical isosceles trihedral angles are congruent. Given two symmetrical isosceles trihedrals and 0' in which L MOP = L M'O'P', L PON = L P'O'N', L NOM = L N'O'M', etc., and L NOM=/L PON and N'O'M' = L P'O'N 1 '. To Prove trihedrals and 0' congruent. Proof. SUG. 1. On account of the symmetry, which face angles and which dihedrals are equal? 2. Because each is isosceles, which face angles are equal? 3. Compare Z NOM with Z P'O'N'; Z PON with Z N'O'M'. 4. From step 3, rearrange the equal parts so that the order is the same for and ' . 507. Therefore 514. VERTICAL POLYHEDRAL ANGLES. When two polyhedrals are so placed that they have a common ver- tex and the sides of the one are extensions through the vertex of the sides of the other, they are vertical poly- hedral angles. 1. Vertical trihedrals are symmetrical. 2. Vertical polyhedrals are symmetrical. 3. What is the locus of points equidistant from the faces of a trihedral? 499. 268 SOLID GEOMETRY 4. What is the greatest number of equilateral triangles which can be put together to form a convex polyhedral angle I 5. How many different sized polyhedrals can be formed from equilateral triangles ? 6. How many different sized polyhedrals can be formed from squares? From .regular pentagons? From regular hexa- gons? 509. 7. How many different polyhedrals can be constructed from regular polygons? 8. Can a mosaic or patch work pattern be constructed from regular triangles? From squares? From regular pentagons? From regular hexagons? From regular heptagons? From regu- lar n-gons? Give reasons for each conclusion. 9. The foundation which bee keepers supply for the bees to build the honey comb upon is a mosaic constructed of regular polygons as near as possible to the shape of a circle. Of what polygons is it formed? 10. A circle would be a better shaped base for the body of the bee. Will the circle or polygon form of construction require the less material? It is assumed in the preceding exercises that the amount of material used in making the cells is proportional to the amount used in making the base. 11. Are two planes perpendicular to the same plane neces- sarily parallel? Are two lines perpendicular to the same plane parallel? Are two planes perpendicular to the same line parallel? 12. Which is the longer, an oblique sect or its projection on a plane? 13. If three non parallel planes are each perpendicular to a fourth plane their three lines of intersection are parallel. 14. If two parallel planes intersect a dihedral, the respective lines of intersection form equal angles. 15. If the mid points of the adjacent sides of a skew quad- rilateral (i. e. one the four vertices of which are not in the same plane) are* joined by straight lines, the figure enclosed is a parallelogram. EXERCISES UGi) 515. TKIRECTANGULAR TRIHEDRAL. A trihedral angle all of the face angles of which are right angles is a tri- rectangular trihedral angle. 16. In a trirectangular trihedral the dihedrals opposite the equal face angles are equal. Why? Is this true of any isosceles trihedral? REVIEW 1 7. State in a theorem a possible condition by which one line can be proved parallel to another; a line can be proved parallel to a plane; a plane be proved parallel to a second plane. 18. State in a theorem a condition by which a line may be proved perpendicular to a second line; to a plane; a plane can be proved perpendicular to a second plane. 19. Stp.te in a theorem a condition by which two trihedrals can be proved congruent ; can be proved symmetrical ; two sym- metrical trihedrals can be proved congruent. 20. What is the locus of points equidistant from the ver- tices of a triangle? 21. Locate a point in a given plane equidistant from all points of a circle which is in another plane. 22. Find a point X which is equidistant from two parallel planes, from two intersecting planes, and from two given points. 23. If two supplementary adjacent dihedrals are bisected by planes, the bisecting planes form a right dihedral. 24. If from any point within a dihedral angle perpendiculars are drawn to the faces of the dihedral, the angle formed is the supplement of the dihedral. 25. If from any point without a dihedral perpendiculars are drawn to the faces of the dihedral, the angle formed by the per- pendiculars is equal to the dihedral. 26. Two trihedrals having two face angles and the included dihedral of the one equal respectively to the corresponding parts of the other are either congruent or symmetrical. 27. Two trihedrals having two dihedrals and the included face angle of the one equal to the corresponding p'arts of the other are either congruent or symmetrical. CHAPTER VIII, POLYHEDRONS. 516. POLYHEDRON. A geometric solid bounded by planes is a. polyhedron. The intersections of the planes are the edges; the intersections of the edges are the ver- tices; the portions of the planes bounded by the edges are the faces; the face upon which the polyhedron is supposed to rest is its base. Any face may be taken as the base. Any straight line connecting two vertices not in the same facets a diagonal of the polyhedron. 517. POLYHEDRONS CLASSIFIED. Polyhedrons are classified according to the number of faces. One of four faces is a tetrahedron, of six faces is a hexahedron, of eight faces is an octohedron, of twelve faces is a dodec- ahedron, of twenty faces is an icosahedron, etc. 1. What is the least number of faces possible for a poly- hedron! 518. LANE SECTION OP A POLYHEDRON. The inter- section of a plane and a polyhedron is a plane section or a section of Mio polyhedron. FE h ;i plmio section of polyhedron. POLYHEDRONS 271 519. CONVEX POLYHEDRON. A polyhedron such that every plane see lion is a convex polygon is a convex poly- hedron. Only convex polyhedrons will be considered in this book. 520. PRISM. A polyhedron bounded by two paral- lel planes and a group of planes the intersections of which are parallel lines is a prism. rP The faces formed by the parallel planes are the bases, usually designated as upper and lower base. The other faces are the lateral faces. The intersections of the lateral faces with each other are the lateral edges and their intersections with the bases are the basal edges. 521. RIGHT SECTION. A section of a prism by a plane perpendicular to the lateral edges is a right sec- tion. 522. OBLIQUE SECTION. A section of a prism by a plane which is oblique to the lateral edges is an oblique section. 523. ALTITUDE OF A PRISM. The distance between the bases of a prism is its altitude. 477. 524. CLASSIFICATION OF PRISMS. Prisms are classified as triangular, quadrangular, etc., according as their bases are triangles, quadrilaterals, etc. PRELIMINARY THEOREMS 525. THEOREM I. The lateral edges of a prism are equal. 479. 526. THEOREM IT. The lateral faces of a prism are parallelograms. 520. 272 SOLID GEOMETRY 527. THEOREM III. The acute angles made by the lateral edges with the planes of the bases are equal. 503. 528. RIGHT PRISM. A prism in which the lateral edges are perpendicular to the bases is a right prism. The edge of a right prism is also its altitude. 529. OBLIQUE PRISM. A prism in which the lateral edges are not perpendicular to the bases is an oblique prism. 530. REGULAR PRISM. A right prism the bases of which are regular polygons is a regular prism. PKOPOSITION I. 531. THEOKEM. Sections of a prism made by parallel planes are congruent polygons. Given ABC and A'B'C', two parallel sections of a prism P. To Prove ABC = A'B'C'. Proof. SUG. 1. What two relations do the lines AB, BC, CD,.... bear to the lines A'B', B'C', C'D', .... respectively ? 478. 2. Compare A ABC, BCD, etc., with A A'B'C', B'C'D', etc., respectively. Auth. Therefore PRISMS 273 532. COR. I. The bases of a prism are congruent polygons. 533. COR. II. A section of a prism made by a plane parallel to th< bas< is congruent to the base. 534. COR. III. All right sections of a prism are congruent polygons. PROPOSITION II. 535. THEOREM. The lateral area of a prism is equal to the product of the perimeter of a right section and a lateral edge. Given the prism M with p the perimeter and a t , a 2 > a v etc., the successive sides of a right section, e the length of the equal lateral edges, and S the lateral area. To Prove S = e x p , Proof. SUG. 1. How do the successive sides of the right section lie with respect to the successive edges which they intersect ? 2. What is the area of the face EF in terms of a t and e ? 3. Express the area of each lateral face. 4. By adding these areas find the total lateral area, 8, in terms of a^ a.,, 3 , etc., and e and then in terms of e and p. Therefore 274 SOLID GEOMETRY 536. COR. The lateral area of a right prism equals the product of tin perimeter of the base and a lateral edge. PROPOSITION III. 537. THEOREM. // two prisms hare the three faces of a trihedral of one congruent to the three faces of a trihedral of the other and similarly placed, the prisms are congruent. E B E' D' Given two prisms P and P' in which the three faces AB, AC, AD forming the trihedral A are respectively congruent to the three faces A'B' ; A'C' , A' D' forming the trihedral A' and are similarly placed. To Prove PEP'. Proof. SUG. 1. Compare trihedrals A and A', 512. 2. Apply P' to P so that face A'B' coincides with face AB. Why can^ this be done ? 3. In what plane does face A'C' fall? FaceA'D'? Why? 511. 4. Where does line A' E' fall? Why.' Where do points E',C',D' fall? Why? Where does plane C'D' fall? Why? 5. Compare faces C'D' and CD. 532. 6. Complete the superposition. 520. Therefore PRISMS 538. COR. I. Two right prisms arc congruent if their altitudes arc equal and their bases congruent. 539. TRUNCATED PRISM. The portion of a prism in- cluded between the base and a section made by a plane not parallel to the base is a truncated prism. 540. COR. II. // two truncated prisms have the flu'cc faces about a trihedral of one congruent respect- ively to the three faces about a trihedral of the other and similarly placed, the truncated prisms are congruent. Proof. Draw figures and make a proof according to the method of 537. PROPOSITION IV. 541. THEOREM. An oblique prism is equal to a right pri.siu the base of which /* a right section of the oblique prism and the altitude of which is equal to the edge of the oblique prism. \ "M L 'D' Given an oblique prism, BC, with a right section A'B' and a lateral edge AC. To Prove prism BC equal to a right prism with base A'B' and an altitude equal to AC. Proof. SUG. 1. Extend the lateral edges making A'C' =AC, and through C' pass a plane I! plane A'B' . Then A' D' is a right prism. 276 SOLID GEOMETRY 2. Compare the faces about the trihe- dral A with the faces about the trihedral C. 3. Compare the truncated prism AB J with the truncated prism CD ' . 4. Compare prism BC with prism B'C'. Therefore 542. PARALLELOPIPEDS. A prism the bases of which are parallelograms is a parallelo- piped. 543. RIGHT PARALLELOPIPED. A parallelepiped the edges of which are perpendicular to the bases is a right parallelepiped. 544. RECTANGULAR PARALLELOPIPED. A right paral- lelopiped the bases of which are rectangles is a rec- tangular parallelepiped. 545. CUBE. A rectangular parallele- piped all the faces of which are squares is a cube. 546. VOLUME OF A POLYHEDRON. The measure of a polyhedron in terms of some other polyhedron taken as the unit of measure is the volume of the polyhedron. 547. UNIT OF MEASURE FOR VOLUME. A cube with an edge equal to a given linear unit is the unit of meas- ure for volume. If a polyhedron contains a cubic inch twenty-five times, the volume of the polyhedron is twenty-five cubic inches. 548. PRELIMINARY THEOREMS. THEOREM T. All faces of a parallelepiped are paral- lelograms. PRISMS 277 THEOREM II. All faces of a rectangular parallelo- piped are rectangles. THEOREM III. All faces of a cube are congruent squares. THEOREM IV. Any pair of opposite faces of a paral- lelopiped man be taken as bases. PROPOSITION V. 549. THEOLEM. Opposite faces of a parallelo- piped are parallel and congruent parallelograms, (ittf/ any section by a plane cutting four parallel edges is a parallelogram. Given a parallelepiped AG with bases AC and EG, AH and EG being one pair of opposite faces, and FHA a section made by a plane cutting four parallel edges. To Prove faces AH and BG parallel and congruent parallelograms, and FD a CU. Proof. SUG. 1. AH and BG are parallelograms. Why? 2. Prove them congruent. 3. Prove them parallel. 478. 4. Prove opposite sides of FD parallel. Therefore 1. Every section of a prism made by a plane parallel to a lateral edge is a parallelogram. 278 SOLID GEOMETRY 1. If from any point iii space perpendiculars are drawn to the lateral faces of a prism, or to the lateral faces extended, these perpendiculars are all in the same plane. 2. Any straight line drawn through the middle point of any diagonal of a parallelopiped, terminating in two opposite faces is bisected at that point. 3. The four diagonals of a rectangular parallelopiped aro equal to one another. 4. The square of a diagonal of a rectangular ptxallelopiped is equal to the sum of the squares on three concurrent edges. 5. The sum of the squares upon the four diagonals of a rect- angular parallelopiped is equal to the sum of the squares upon the twelve edges. 6. Prove the preceding exercise for any parallelopiped. S8 337, 338. 7. Prove that the lateral area of a right prism is less than the lateral area of any oblique prism having the same base and an equal altitude. Sue. Draw an oblique and a right prism upon the game base with equal altitudes. No face of the oblique prism has a less altitude than the corresponding face of the right prism. Why ? Some faces of the oblique prism must have altitudes greater than those of the correspond- ing faces of the right prism. Why? The altitudes of the faces are defined with respect to the common bases. 8. Make a list of theorems on polyhedrals and in connec- tion with each write out a plane geometry theorem which closely corresponds to it. 9. From two points on the same side of a plane draw two lines to a point in the plane that shall make equal angles with the plane. 10. The lines joining the mid points of opposite sides of a skew quadrilateral bisect each other. Ex. 11. A plane perpendicular to a line in a plane is perpendicu lar to that plane. 12. If two or more planes intersect in one straight line, the perpendiculars drawn to them from any one point lie in the same plane. PRISMS 279 PROPOSITION VI. 550. THEOREM. A plane passed thro null tico diagonally opposite edges of a parallelepiped di- vides the parallelepiped Into two equal triangular prisms. D' B Given parallelepipeds CD' divided by a plane through two opposite edges CC' and DD' into two triangular prisms A'B'C'- A and B'C'D'-B. To Prove A'B'C'- A = B'C'D'- B. Proof. SUG. 1. A right section as HF intersects a CH on the parallelepiped and this is divided into two congruent A by the plane CD'. Why? 2. Compare A-prism A'C'D' A with a right prism on EHF as base with an altitude equal to D'D. Compare A-prism B'C'D' B with a right prism on GHF as base with an alti- tude equal to C'C. Auth. 3. Compare these two right prisms. 538. 4. Compare A'C'D' -A with B'C'D'-B. Therefore In the above figure which prisms are congruent and which are only equal? 1. In 550 prove A'B'C' A and B'C'D' B symmetrical. 2. The four diagonals of a parallelepiped bisect each other. 280 SOLID GEOMETRY PROPOSITION VII. 551. THEOREM. Two rectangular parallelo- pipeds having congruent bases are proportional to their altitudes. Given two rectangular parallelepipeds P and P' with congruent bases and altitudes a and a' respectively. P a To Prove = . P' a Proof. Case I. a and a' commensurable. SUG. 1. Divide the altitudes by a common unit of measure, supposing it to be contained m and m' times in a and a' respectively. What is the ratio of the altitudes? 2. Through the points of division of the altitudes pass planes parallel to the bases. What kind of parallelepipeds are formed? Compare them in number and volume. Auth. What is the ratio of P and P' ? 3. Compare the ratios of the altitudes and the parallelepipeds. Case II. a and a' incommensurable. SUG. 1. Divide the altitudes a and a' by any unit commensurable with a'. In the division of a there will be a remainder x less than the unit. Why? By taking this unit smaller and smaller PRISMS 281 this remainder may be made to decrease indefi- nitely. Why ? M 1 2. Through this last point of division, M, pass a plane parallel to the base. As the unit in use is decreased what change takes place in the altitude 031? What change takes place in the parallelepiped MN1 In the parallelepiped with altitude xt What is the limit of the varia- ble altitude OMf Of the variable parallele- piped MA 7 ? 0.17 .a . MN . P 3. -- = and - a' a P' P' MX OM Wl 4. == . Why? P' a 5. ' = -. 426. P' a Therefore 552. DIMENSIONS OF A PARALLELOPIPED. The dimen- sions of a parallelepiped are its altitude and the base c^nd altitude of its base. Consequently the three dimen- sions of a rectangular parallelepiped are any three con- current edges. 282 SOLID GEOMETRY PROPOSITION VIII. 553. THEOREM. Two rectangular parallelepi- peds having equal altitudes are proportional to iheir bases. \ P' Given two rectangular parallelepipeds P and P' with equal altitudes a, their bases B and B' having the di- mensions b, c and b' , c' respectively. P B To Prove = Proof. SUG. 1. Construct a third rectangular par- allelepiped N with the altitude a and base with dimensions b' ,c. 2. P b , N c - = and = N b' P' .,,., V, hy ? 3. Hence - = c - = -. Why? P' ])'c B' Therefore 1. A gable for a dormer window is triangular in shape with an angle of 60 at the ridge. If the length of the rafters is 6 ft., what is the area of the largest circular window that can be in- serted, making allowance for a four inch frame around the win- dow! 2. The diagonals of a rectangular parallelepiped are equal. 3. Determine a point in one side of a triangle equally from the other two sides. PRISMS 283 PROPOSITION IX. 554. THEOREM. Two rectangular parallelo- pipeds are to each other as the product of their three dimensions. I Given two rectangular parallelepipeds P and P', with dimensions a, b, c and a', b' , c' respectively. P axbxc To Prove P' a'xb'xc' Proof. SUG. 1. Let N be a rectangular parallele- piped with the dimensions a, b, c' . j axb ,. 2. Then = - and = - W hy ? PC jN axb = - and = - N c' P' a'xb' 3. -=-. Why? P' a'xb'xc' Therefore 555. COR. I. The volume of a rectangular parallele- piped is the product of its three dimension*. Given a rectangular parallelepiped P with dimensions a, 6, c. To Prove Vol. P = abc. Proof. SUG. 1. Take as the unit of volume a cube U with an edge equal to the linear unit in which a, 6, c are expressed. 284 SOLID GEOMETRY i P axbxc 2. - = abc. U Ixlxl 3. ' Vol. P = abc. U. Therefore In the applications of this theorem the three dimensions must be expressed in terms of the same linear unit and the unit of volume must be a cube with an edge equal to this linear unit. 556. COR. Two rectangular parallelepipeds having two dimensions respectively equal are to each other as their third dimensions. 557. COR. II. The volume of a cube equals the cube of its edge. 558. By comparison of the theorem and the defini- tion of volume (546) it will be observed that the vol- ume of a rectangular parallelepiped is equal to the product of the measures of the three edges meeting at any vertex times the unit of volume. The expression ' ' product of the three dimensions ' ' is un- derstood to mean the product of the mea- sures of these lines. 361 note. When each dimension of the rectangu- lar parallelepiped is divisible by the linear unit which is the edge of the unit vol- ume, the truth of the theorem on volume may be shown by dividing the parallelepiped into unit cubes. 1. What is the volume of a cube the edge of which is 7 in.? 2. What is the volume of a rectangular parallelepiped the dimensions of which are 4 in., 7 in., 12 in.? 6 ft. VI 2 ft., \/TS" ft.? VlT, \'T8, V24? V87V24, vT2~? 3. The edges of a rectangular parallelopiped are 8, 12, 16. What is the length of ;i diagonal? What is its length if Ilio dimensions are V&7 v!>. \'lx> 4. Find the lot:! I surface area of a regular triangular prism with basal edges of 4 ft. and latorn! edges of x ft. PRISMS 285 PROPOSITION X. 559. THEOREM. Any parallelepiped is equal t(> a rectangular parallelopiped having an equal base and altitude. -L'' / B'^ / t - --/-, / Given a parallelopiped P with base ABC and alti- tude h. To Prove P a rectangular parallelopiped H having a base EFG = ABC and altitude h. Proof. SUG. 1. Extend edge BC of P and the three edges parallel to BC and at some convenient point on line BC take B'C' =BC. Through B' and C" pass planes perpendicular to line B'C', forming the parallelopiped X with base A' B'C'. Extend the edge A 'B' and the three edges parallel to A'B' and at some convenieut point take EF = A'B'. Through E and F pass planes per- pendicular to EF forming the parallelopiped H. 2. Show that N is a right parallelo- 286 SOLID GEOMETRY piped and H is a rectangular parallelepiped. Show that P, N, and H have the same altitude and equal bases. 3. Compare P and N. 548 IV, 541. Compare N and H. 4. Compare P and H. Therefore 560. COR. I. The volume of any parallelopiped is equal to the product of its three dimensions. Given a parallelopiped P with dimensions a, b, c. To Prove Vol. P = abc. Proof. P equals a rectangular parallelopiped with dimensions a, b, c. Why? The volume of this second parallelopiped is abc. Therefore 561. COR. II. Two parallelopipeds u'ith equal bases are to each other as their altitudes. Proof. Let the dimensions be a, b, c and a, b, c' , respectively. Use 560 to find the ratio. 562. COR. III. Two parallelopipeds with equal alti- tudes are to each other as their bases. Proof. Let the dimensions be a, b, c, and a, b ' , c' respectively. Use 560 to find the ratio. 563. COR. IV. Two parallelopipeds are to each other as the products of their three dimensions. 1. Find the lateral area of a regular pentagonal prism each edge of which is 3 in. 2. If a secant plane intersects two planes so that the lines of intersection are parallel and the corresponding dihedrals are equal, the two planes are parallel. 3. Prove the preceding example when the equal dihedrals are the alternate interior angles; prove it for equal alternate ex- terior angles. PRISMS 287 PROPOSITION XI. 564. THEOREM. The volume of a triangular prism is equal to the product of its base and its altitude. F Given the triangular prism EFG N, denoting its volume by V, its base by B, and its altitude by h. To Prove V = hB. Proof. SUG. 1. Extend the planes of the bases. Through the edges AE and CG pass planes paral- lel to the faces NG and NE respectively. The figure DF is a parallelepiped. Why ? 2. What is the volume of DF in terms of B and h / Why? 3. What is the volume of the A-prism in terms of DF ? 550. 4. What is the volume of the A-prism in terms of B. and h ? Therefore 1. Compare the volumes of two rectangular parallelepipeds the respective edges of which are 2', 3', 7' and 5', 3/, 8'. 2. A parallelepiped has an altitude of 8 in. and its base is a, rhombus with a 10 in. side, the shorter diagonal being 12 in. Find the volume. 288 SOLID GEOMETRY PROPOSITION XII. 565. THEOREM. The volume of any prism is equal to the product of its base and its attitude. Given the prism GD with V, B, and h denoting its volume, base, and altitude respectively. To Prove V = hB. Proof. SUG. 1. Through any one lateral edge pass diagonal planes. Into what kind of figures is the prism divided? 2. Denote the respective volumes and bases of these figures by V lf B l ; V 2 , B F etc. Then 8 , V 3 = h X B 3 , etc. Why? 3. V=V l +V 2 +V 8 .... 4. Determine V in terms of h and B. Therefore 566. COR. I. // two prisms have equal bases, their volumes are proportional to their altitudes. 567. COR. II. // two prisms have equal altitudes their volumes are proportional to their bases. EXERCISES 280 1. Find the volume of a regular hexagonal prism, the lateral edge being 10 in. and the basal edge being 5 in. 2. Find the volume of a rectangular parallelepiped, the lateral edge being 20' and the basal edges being 5'. 3. Find the volume of a parallelepiped, the base being 8" square, the lateral edge 13" and the perpendicular let fall from a vertex of one base striking the other base 5" from the corre- sponding vertex. 4. Find the volume of a hexagonal prism, having for its base a regular polygon, the lateral edge being 25", the basal edge 10", and the inclination of the lateral edge to the base be- ing 60. Find the volume if the inclination is 45. 5. The specific gravity of iron is 7. 4. What is the weight of a rectangular tank including the cover \" thick, made of iron the inside measurements being 20" X 20" X 4'? Make a diagram showing how the answer can be obtained with the least computation. 6. A right triangular tank has a lateral edge of 15' and basal edges of 8', 7', and 5', inside measurements. Find its total area and its capacity in gallons. 7J gal. = 1 cu. ft. approximately. 7. How can one obtain the volume of an irregular piece of rock by immersing it? 8. What is the ratio of two rectangular solids the dimen- sions of which are 3, 8, 12 and 4, 9, 20 respectively? 9. The apothem of a regular hexagonal prism is 10 and its lateral edge is 20. Find its volume and total area. 10. The apothem of a cube is 4. Find volume and total area. 11. The edges of three cubes are 7", 12", and 13" respectively. Find the volume of each and the total area of each. 12. A rectangular box 12" X 18" X 22", outside measure- ment, is made of 1" boards. What is its capacity in cubic inches? How many cubical boxes 2f" on an edge can be packed in it? What is its capacity in gallons? 13. A cistern is in the form of a regular hexagonal prism. The lateral edge is 7' and the basal edge is 6', inside measure- ments. What is its capacity in gallons? 14. The volume of a rectangular parallelepiped is 6,720 cubic inches and its edges are in the ratio of 3, 5, 7. Find the three 290 SOLID GEOMETRY 568. PYRAMID. A polyhedron all but one of the faces of which meet in the same point is a pyramid. The point in which these faces meet is the vertex. The face which does not meet the vertex is the base and the other faces are the lateral faces. The intersections of the lateral faces are the lateral edges D < and the intersections of the lateral faces with the base are the basal edges. Point out the various parts of the pyramid above. 569. ALTITUDE OF A PYRAMID. The length of the perpendicular from the vertex to the plane of the base is the altitude of the pyramid. It may also be taken as the perpendicular distance between the plane of the base and a plane through the vertex parallel to the base. Why? 570. PYRAMIDS CLASSIFIED. A pyramid is triangu- lar, quadrangular, pentagonal, etc., according as its base is a triangle, quadrilateral, a pentagon, etc. In a triangular pyramid any face may be taken for the base, the vertex of the opposite polyhedral angle then being the ver- tex of the pyramid. 571. REGULAR PYRAMID. A pyramid with a regular base, such that the vertex lies in the perpendicular erected at the center of the base, is a regular pyramid. 572. SLANT HEIGHT. The perpendicular from the vertex of a regular pyramid to any basal edge is the slant height of the regular pyramid. 573. TRUNCATED PYRAMID. That portion of a pyra- mid included between the base and a plane cutting all the lateral edges is a truncated pyramid. 574. FRUSTUM OF A PYRAMID. A truncated pyramid PYRAMIDS 291 iii which the cutting plane is parallel to the base is a frustum of a pyramid. The section made by the cutting plane is the upper base of the frustum. 575. ALTITUDE OF A FRUSTUM. The perpendicular distance between the bases is the altitude of the frus- tum. COROLLARIES TO THE DEFINITIONS. 576. COR. I. The lateral faces of a pyramid are tri- angles. 577. COR. II. In a regular pyramid the lateral edges are equal, the lateral faces are congruent triangles, arid the slant height is the same irrespective of the face in uHiich it is drawn. 578. COR. III. In a frustum of a regular pyramid the lateral edges are equal, the lateral faces are con- gruent trapezoids, and the slant height is the same irre- spective of the face in ivhich it is drawn. PROPOSITION XIII. 579. THEOREM. // a pyramid is cut by a plane parallel to the base, I. The lateral edges and the altitude are cut proportionally; 292 SOLID GEOMETRY II. The section is a polygon similar to the base. Given a pyramid P AC with base ABC... cut by a plane M II the base in the section A'B'C' . . . with alti- tude PO. _ _ PA PB PC PO I. To Prove = = , etc.. = PA' PB' 'PC' PO' Proof. SUG. 1. Through the vertex P pass plane N II plane M. 2. Compare the ratios PA PB PC PO 9 ' P, i. e. P' P equals some definite number K. By 586 P' < V and P > V so that P'- P < V'-V. 49. 5. Hence P' P < A' . Why? By PYRAMIDS 297 taking the length x small enough the prism A' can be made as small as is desired, even less than K, since its altitude x is decreased. 564. Hence P' cannot be greater than P. 6. Form the inscribed prisms in P' and the circumscribed prisms in P. It can now be proved that P' cannot be less than P. Therefore PROPOSITION XIX. 588. THEOREM. The volume of a triangular pyramid is one-third the product of its base and its altitude. A n- 7 Given the triangular pyramid A ECD, its volume denoted by V, its base by B, and its altitude by h. To Prove V= $hxB. Proof. SUG. 1. Through A pass a plane parallel to the base, extend the planes of the faces ACE and ACD, and through ED pass a plane parallel to AC. The resulting figure is a prism. Why? 2. A EE'D'D is a quadrangular The plane of AE'D divides it into two pyramids, A EE'D and A E'D'D. 3. AECD = DE'D'A = AE'DE 4. A ECD = J prism. What is its base and altitude ? Authority for each statement. prism, equal 298 SOLID GEOMETRY 6. Complete the demonstration. Therefore PROPOSITION XX. 589. THEOREM. The volume of any pyramid is equal to the product of its base and its altitude. Given pyramid 0, denoting its volume by I 7 , its base by B, and its altitude by h. To Prove V = J h x B. Proof. The proof, similar to that of 565, is left to the pupil. 590. COR. I. // two pyramids have equal bases then- volumes have the same ratio as their altitudes. 591. COB. II. // two pyramids have the same alti- tude, their volumes have the same ratio as their bases. COR. III. Any two pyramids are proportional to the products of their bases and altitudes. PROPOSITION XXI. 592. THEOREM. The volume of a frustum of a triangular pyramid is equal to the sum of the vol- umes of three triangular pyramids each with the altitude of the frustum and with bases equal rr PYRAMIDS 299 spectively to the upper base of the frustum, the lower base of the frustum, and a mean propor- tional to the bases of the frustum. Given frustum of a triangular pyramid F, with upper base B l and lower base B 2 . To Prove F = $ Ji(B l + B 2 + ^B~B~ 2 ). Proof. SUG. 1. By a plane through GE'F' cut off a triangular pyramid P, with base B l and altitude h. What is its volume ? 2. By a plane through GEF' cut off a triangular pyramid P 8 with base B% and alti- tude h. What is its volume ? 3. The remaining portion is a triangu- lar pyramid, P 3 . A/H^__^fJ^_ Vff^ Wh . ? /xt^ff. f," /n " rv Why? &GE'G' G'E Therefore 300 SOLID GEOMETRY PROPOSITION XXII. 593. THEOREM. The volume of the frustum of any pyramid is equal to the sum of the volumes of three pyramids each with the altitude of the frustum and with bases equal respectively to the upper base of the frustum, the lower base of the frustum, and a mean proportional to the two bases of the frustum. Given the frustum F with upper base B 19 lower base B 2 , and altitude h. To Prove F = % h (J? x + 5 8 + vB~B~ 2 ). Proof. SUG. 1. The lateral edges of F will if ex- tended, meet in a point forming a pyramid P. Why? Construct a triangular pyramid P' with the same altitude a as P and a base B 2 ' equal to the base B 2 of P. Cut from P f a frustum F' with altitude- h. ' 2. Compare the volumes of P and P'. Auth. 3. Compare #/ and ,. 582. 4. The pyramids with bases B^ and B/and altitudes a ft are equal. Why? PYRAMIDS 301 5. Hence the frustum F = F'. Why? 6. F'=$h(B 1 '+B 2 '+^B~'B./) Why? Therefore 1. A monument is 25' high, 18" square at one end, 30" square at the other, and of uniform shape. What is its volume in cubic feet? If its specific gravity is 7-J what does it weigh in tons? PBOPOSITION XXIII. 594. THEOREM. Two tetrahedrons having a trihedral angle of one equal to a trihedral angle of the other have the same ratio as the products of the three edges including the equal trihedral angles. C' Given two tetrahedrons ABC and 0' A'B'C' with equal trihedrals and 0'. To Prove ~ ABC - ^xQJ8xOC 0' -A'B'C' O'A'xO'B'xO'C' Proof. SUG. 1. Superimpose trihedral 0' upon tri- hedral and from points A and A ' drop perpen- diculars to the opposite face meeting it in points M and M' respectively. 2 Then ~ ABC = AM x A OBC 0- A'B'C' A'M'x&OB'C'' Why? 302 SOLID GEOMETRY t &OBC OBxOC 3. Also - = . Why? AOB'C" OB'xOC' 4. Points 0, M, M' are in a straight line. Why ? 5 ,AK = oA_. Why? A'M' OA' 6. Complete the demonstration. Therefore 595. SIMILAR POLYHEDRONS. If two polyhedrons have the same number of faces similar each to each and similarly placed and have their corresponding polyhe- dral angles equal, the polyhedrons are similar. The equal angles and the lines and faces in the two polyhedrons which are similarly situated are called homologous angles, lines, and faces, respectively. PRELIMINARY THEOREMS. 596. THEOREM I. Homologous lines in similar poly- hedrons are proportional. 597. THEOREM II. Homologous faces in similar polyhedrons are proportional to the squares of homo- logous lines. 598. THEOREM III. The surfaces of similar poly- hedrons are proportional to the squares of homologous lines. 374. 1. Find the volume of a regular triangular pyramid with basal edge of 4 ft. and altitude of 5 V~3 ft. 2. The point of meeting of the three medians of an equi- lateral triangle is f of the length of the median from each ver- tex. Ex. 30, p. 212. 3. The edges of a regular tetrahedron are each a ft. Find its slant height, altitude, base area, lateral area, total area, and vc lume. 4. Find the volume of a regular hexagonal prism with a radius of 2 ft. and lateral edge of 2 yards. PYRAMIDS 303 PROPOSITION XXIV. 599. THEOREM. Two similar tetrahedrons are proportional to the cubes of homologous edges. O' Given two similar tetrahedrons, and 0' , edges OA, OB, OC ... being homologous to O'A', O'B', O'C/ . . respectively. To Prove =M. 0' OA'* Proof. SUG. 1. Compare the trihedral angles and Q: - = ^.x-.X--. Why 0' O'A'xO'B'xO'C' O'A' O'B' O'C' f) R Of 1 3. Substitute for - -and- - in (1), O'B' O'C 1. Volumes of similar solids are to each other as the square roots of the cubes of their surfaces. 2. If the amount of lumber used is to be the same in either case which provides the greater capacity, one barn or two, the barns being similar in shape? 3. If the dimensions of a window be doubled, by what fac- tor is its lighting capacity increased? 4. If the surface of a box be trebled in building a second box of the same shape, by what is the capacity Increased? 5. What increase in material is required in building a new grain box having the same shape as the old one but of double the capacity? 304 SOLID GEOMETRY PROPOSITION XXV. 600. THEOREM. Two tetrahedrons are simi- lar if three faces of one are respectively similar to three faces of the other. ir, C' Given two tetrahedrons and 0' with the three faces about similar to the three faces about ' respectively. To Prove O^O'. Compare trihedrals and 0'. 1. Proof. SUG. Auth. 2. A'B', B'C' with 3. Compare lines AB, BC, CA C' A' respectively. Auth. Compare & ABC and A'B'C'. Auth. 4. The corresponding trihedrals and dihedrals are equal. Why? 5. The conditions for similarity are fulfilled. Why? Therefore 601. COB. Two tetrahedrons having a dihedral of one equal to a dihedral of the other and the including faces of the first dihedral similar respectively to the in- cluding faces of the second dihedral and similarly placed are similar. Proof. SUG. Prove the faces opposite the equal dihedrals similar. PYRAMIDS PROPOSITION XXVI. 305 602. THEOREM. Two similar polyhedrons can be divided into tetrahedrons similar each to each and similarly placed. Given P and P' similar polyhedrons. To Prove that P and P' can be divided into tetra- hedrons similar each, to each and similarly placed. Proof. SUG. 1. Through any two corresponding vertices as and 0' draw in each of the includ- ing faces all possible diagonals. The correspond- ing faces of P and P' are thus divided into the same number of triangles, similar each to each and similarly placed. Why? 2. Let OE and O'E' be two corre- sponding edges and let OF and OG be the two diagonals including between them OE. Simi- larly let O'F' and O'G' be the corresponding lines of P'. Cut from P and P' the respective tetrahedrons EFG and 0' E'F'G'. By Sug. (1) kOEF'-' kO'E'F' and A OEG <-' A O'E'G'. Also dihedral OE = dihedral O'E'. Why? Hence tetrahedron EFG ^ tetrahedron 0' E'F'G'. 601. 3. In the remaining portions of P and P' the new face A FOG and A EFG are similar respectively to F'O'G' and E'F'G'. Why? 306 SOLID GEOMETRY Also the ratio of the new edges OF, OG, FG to O'F', O'G', F'G' respectively, being equal to OF - ; (Why?) the original ratio of similitude. O'E' 4. Since the trihedral angles and ' of the removed tetrahedrons are equal, so are the remaining polyhedral angles and 0' . 5. Hence the remaining polyhedrals are similar. Why? 6. Repeat the process on the remaining portions of P and P' . Therefore 1. Similar polyhedrons have the same ratio as the cubes of homologous edges. Sue. According to the demonstration of 602, the poly- hedrons may be divided into tetrahedrons. Let them be respectively P t P/, P 2 P 2 ' etc. Let a t and a/ be homol ogous edges of P a and PI', etc. Then by 599. ^ = *, > *- f =. -&J-, etc. But the ratios of the a 's are "i a s PZ a z 3 equal. Why! Hence the ratios of the P's are equal and P t + P 2 + P a + . . . _ Pi_ _ a^. P/+ P 2 '+ P 3 '+ . . . P/ fll ' 603. REGULAR POLYHEDRON. A polyhedron in whicli all the faces are regular congruent polygons and in which the polyhedral angles are equal is a regular poly- hedron. A regular polyhedron must be convex. 2. The perimeter of the mid section of a frustum of a pyra- mid equals one-half the sum of the perimeters of the bases. 3. A grain bin holds 100 bu. Another bin has each dimen- sion equal to twice the corresponding dimension of the first. What is its capacity? 4. The surfaces of two similar solids are to each other as the cube roots of the squares of their volumes. REGULAR POLYHEDRONS 307 PROPOSITION XXVII. 604. THEOREM. At most only five regular polyhedrons can be formed. Proof. SUG. 1. Show that three, four, or five regu- lar triangles can be so used as to form a convex polyhedral angle. 2. Show that more than five regular triangles cannot form a convex polyhedral. 3. Make a similar test for the regular tetragon or square. 510. 4. How many regular pentagons may be used? Auth. 5. Show that no regular polygon of more than five sides can be used to form a convex polyhedral. Therefore 605. SCHOLIUM. That five regular polygons can all be constructed is illustrated as follows : Cut the patterns below from cardboard and cut the material half through along the dotted lines. Fold each so as to form a poly- hedron, pasting strips of paper along the edges to keep them in shape. It is possible to prove that these figures can actually exist by purely mathematical reasoning. 308 SOLID GEOMETRY i 1. The total areas of regular polyhedrons have the same ratios as the squares of their altitudes or as the squares of any two homologous edges. 2. Find the volume of a regular quadrangular pyramid with basal edge of 8' and slant height of 5'. 3. The volume of a truncated triangular prism is equal to the sum of the volumes of three pyramids having the base of the prism as a common base and the vertices of the inclined section as respective vertices. To PROVE EFD ABC = E ABC + V ABC + F ABC. Pass planes through each vertex of the truncating section, DEF, and the respective opposite edges of the base. Plane EAC cuts off the first pyramid. There is left E-^ADCF. Plane EDC cuts off E ACD = B ABC oi-D ABC, 570 the second pyramid. The remaining pyramid, E DCF or D FEC 1) FCB (equal bases). But D FCB = AFCB, the third pyramid 4. Find the volume of a truncated triangular right prism with basal edges of 5', 5', 8' and lateral edges of 7', 8', 9'respeet- ively. Use two methods of finding the base. 5. The volume of any truncated triangular prism equals the product of a right section and one-third the sum of the lateral edges. 6. The sides of a right section of a truncated triangular pyramid are 8', 9', 15' and the lateral edges are 5', 9', 32' respect- ively. Find the lateral area and the volume. 7. The apothem of a cube is 7'. Find the volume by using (he rule for pyramids. 8. Find the volume of a monolith '24' high, J)' square at one end and 4' square at the other. Solve without pencil. 9. Find the slant height, the apothem, the lateral surl'.-icc, the total surface, and the volume of a regular tetrahedron with an edge of 5. CHAPTER IX. THE THREE ROUND BODIES. 606. CYLINDRICAL SURFACE. A surface formed by a moving straight line which always remains parallel to its original position and continually touches a fixed curved line is a cylindrical surface. The moving straight line is the generatrix and the fixed curve is the directrix. The generatrix in any one of its posi- tions is an element of the surface. If the directrix is a closed curve, the cylindrical surface is a closed cylindrical surface. 607. A CYLINDER. A solid bounded by a closed cylin- drical surface and two parallel planes cutting the ele- ments is a cylinder. The plane sur- faces are the bases of the cylinder and the cylindrical surface is the lateral surface of the cylinder. The distance between the two bases is the altitude of the cvlinder. - 608. RIGHT SECTION. A section of a cylinder made by a plane perpendicular to an element is a right section. 609. CIRCULAR CYLINDER. A cylinder the bases of which are circles is a circular cylinder. The line joining the centers of the bases is the axis of the cylinder. 310 SOLID GEOMETRY i 610. RIGHT CYLINDER. A cylinder the bases of which are perpendicular to the elements is a right cylinder. 611. OBLIQUE CYLINDER. A cylinder in which the elements are not perpendicular to the bases is an oblique cylinder. 612. CYLINDER OF REVOLUTION. A cylinder generated by the revolution of a rectangle about one of its sides is a cylinder of revolution. 613. SIMILAR CYLINDERS. Cylinders generated by the revolution of similar rectangles about homologous sides are similar cylinders. 614. TANGENTS TO A CYLINDER. If a line touches the lateral surface of a cylinder but does not intersect it, it is tangent to the cylinder. If a plane embraces an element of a cylinder but does not intersect the surface, it is tangent to the cyl- inder. EF is tangent to the cylinder at and plane N is tangent in the element GH. PRELIMINARY THEOREMS. 615. THEOREM I. The elements of a cylinder are parallel and equal. 616. THEOREM II. A right section of a cylinder is perpendicular to every element. 617. THEOREM III. A cylinder of revolution is a right circular cylinder. 1. Sect a is perpendicular to sort ?>. Tf ft is revolved about its free extremity in one plane what figure is formed by a .' CYLINDERS 311 PROPOSITION I. 618. THEOREM. Every section of a cylinder made by a plane embracing an element is a par- allelogram. Given the plane AD embracing the element AB. To Prove that plane AD intersects the surface of the cylinder in ZZ7 ABCD. Proof. SUG. 1. Let D be the point in which the plane cuts the perimeter of the base the second time and through D draw the element DC. How does DC lie with respect to BAt Why? 2. DC and BA determine a plane con- taining BA and D. "Why ? 3. How many planes can contain BA and D ? Where then does DC lie with respect to the given plane ABDt With respect to the in- tersection of the plane and the cylinder? 4. Complete the proof by showing that Therefore 619. COR. A plane containing an element of a cylin- drical surface without being tangent inter seels the sur- face in a second element also. 620. COR. II. Every section of a right cylinder by a plane which embraces an element is a rectangle. ]. Cut a cylinder of revolution by a plane parallel to an element in such a manner that the section shall be a rectangle congruent to the rectangle which generates the cylinder. 312 SOLID GEOMETRY . PROPOSITION II. 621. THEOREM. The bases of a cylinder are congruent. Given a cylinder AD with bases ABC and DEF. To Prove bases ABC and DEF congruent. Proof. SUG. 1. Let A, B, C be any three points in the perimeter of one base and draw the element AD. Pass planes through the element AD and the points B and (7, intersecting the cylinder in the elements BE and CF respectively. ADEB, ADFC, BCFE are /I7. Why? 2. Compare & ABC and DEF. Auth. 3. Superimpose A ABC upon A DEF. As A, B, and C are any three points of the perim- eter ABC, where does the perimeter ABC fall? 4. Compare the two bases. Therefore 622. COR. I. Any two parallel sections cutting the elements of a cylinder are congruent. 623. COR. II. All sections of a cylinder parallel to the bases are congruent to the bases. 624. COR. III. The axis of a circular cylinder passes through the centers of all sections which are parallel to the CYLINDERS 313 SUG. Draw any two diameters in one base. Pass planes through these two diameters and the ele- ments at their extremities. Where will these two planes intersect the section? The other base? Where does the axis lie with respect to these planes ? 1. A plane tangent to a cylinder of revolution is perpen- dicular to the plane of any right section. 625. CYLINDER INSCRIBED IN A PRISM. If each lateral face of a prism is tangent to a cylinder and the bases of the prism are circum- scribed about the corresponding bases of the cylinder, the prism is circumscribed about the cylinder and the cylinder is inscribed in the prism. 626. CYLINDER CIRCUMSCRIBED ABOUT A PRISM. If each lateral edge of a prism is an element of a cylinder and the bases of the prism are inscribed in the corresponding bases of the cylinder, the prism is inscribed in the cylinder and the cylinder is circumscribed about the prism. 627. POSTULATE I. The volume of a circumscribed prism, the areas and perimeters of its bases and sections, and its lateral area are greater than the corresponding parts of an inscribed cylinder. 314 SOLID GEOMETKY 628. POSTULATE II. The volume of an inscribed prism, the areas and perimeters of its bases and sections, and its lateral area are less than the corresponding parts of a circumscribed cylinder. 629. POSTULATE III. // the number of sides of a prism inscribed in a cylinder or circumscribed about a cylinder be indefinitely increased in such a manner that the sides of the prism be all indefinitely decreased, the volume, lateral surface, bases, sections, perimeters of bases and sections of the cylinder are the limits of ihe corresponding parts of the prism. PROPOSITION III. 630. THEOREM. The area of the lateral sur- face of a cylinder is equal to the perimeter of a right section multiplied by an element of the sur- face. Given cylinder AG, its lateral area denoted by S, the perimeter of the right section by p, and the element AE by e. To Prove S = p x e , Proof. SUG. 1. Inscribe in the cylinder a prism, d3noting its lateral area by S', the perimeter of its right section p by p'. Its edge is e. Why? 2. Then S'=p' X e. Why? 3. Indefinitely increase the number of CYLINDERS 315 faces in the manner indicated in 629. What are the limits of S', p' , and p' X el 4. 'S = p*e. Why? Therefore 631. COB. I. The lateral area of a cylinder of revo- lution is equal to the product of the circumference of the base and an element, or the altitude. i. e. S = 2"rh. 632. COR. II. The lateral areas of similar cylinders are to each other as the squares of their altitudes and as the squares of the radii (or diameters) of their bases. SUG. Denote the lateral areas by 8 and S 2 , the altitudes by h^ and h 2 , the radii by r i and r 2 , the diameters by d l and <2 2 respectively. Then ^i = 27r ^ fti = *i fti = r, x fei = r, 2 = fe, == d 1 ^ # 2 2*r 2 7i 2 r 2 7i 2 r 2 7i 2 r 2 2 7i 2 2 d 2 2 ' Give the reasons for each statement. 633. COR. III. The total area, A, of a cylinder of revolution, which is the sum of the lateral area and the areas of the two bases, is given by the formula A = 2*rh 634. COR. IV. The total areas of two similar cylin- ders are to each other as the squares of their altitudes, and as the squares of their radii (or diameters). Proof left to the student. 1. How many feet of lumber will it take to build the walls of a circular silo 15' in diameter and 25' high, allowing ^ for matching? 2. A cistern is 10' deep and 8' in diameter. How many square yards of cement is needed to line it? 3. The total area of a right circular cylinder is SOvr sq. ft. and the radius of the base is 5 ft. Find the altitude. 4. Which generates the greater lateral area, the revolution of a rectangle about its longer side or about its shorter side? 316 SOLID GEOMETRY t PROPOSITION IV. 635. THEOREM. The volume of a cylinder is equal to the product of its base and its altitude. Given a cylinder Fft, its volume denoted by V, its base by B, and its altiude by h. To Prove V = B*h. Proof. SUG. 1. Inscribe in the cylinder a prism, its volume denoted by V and its base by B' . Its altitude will be h. Why ? 2. Then V'=B' x ft. Why? 3. Indefinitely increase the number of sides of the prism in the manner indicated in 629. What is the limit of V ? Of IT ? Of the product fl'Xft? Why? 4. Complete the demonstration. Therefore 636 COR. I. The volume of a cylinder of revolution is expressed in terms of its altitude h and radius r by the formula V = 7r r 2 h. 637. COR. II. The volumes of two similar cylinders of revolution are to each other as the cubes of their alti- tudes and as the cubes of their radii (or their diam- eters). Proof loft to the pupil. See 632. EXERCISES 317 1. What must be the shape of a piece of paper which ex- actly covers the lateral surface of an oblique cylinder? Of a right cylinder? 2. A cistern holding 75 bbls. is 8 ft. deep. What is its diameter? How many square yards of cement are required to line it? 3. A cylindrical tank on a water works tower has a lateral area of 1,232 sq. ft. The radius of its base is \ the altitude. Find the altitude and the radius. Find the total area. What would be the lateral area if the altitude were doubled. Find its total area under the same condition. 4. What is the locus of a point in space at a given distance from an unlimited straight line? 5. A saw log is 16' long and 18" in diameter at the small end. How many board feet in the largest squared timber that tan be sawed from it? 6. A cylindrical water reservoir is 110' in diameter and 20' deep. How many bbls. does it hold? How many square yards of cement are necessary to line it? 7. A rock submerged in a tank 29' in diameter and 8' deep raised the water 2'. What was the volume of the rock? 8. A farmer feeds his 60 cows 2 bu. of ensilage each a day. The silo is cylindrical, 30' in diameter and 40' deep. How long will the ensilage last? 0. There are wine tanks 15' deep and 10' in diameter. How many quart bottles can be filled from one of them? 10. What is the ratio of the lateral area of a right circular cylinder to the sum of the bases? 11. What is the locus of a sect at a given distance from a straight line to which it is parallel? 12. What is the locus of point X which is at a given distance from a straight line and equally distant from two fixed points? 13. What is the locus of a point at a given distance from the lateral surface of a cylinder? Is it possible for one element of the locus to lie within the cylinder? Under what condition? 14. From a given point without a cylinder draw a plane tan gent to the cylinder. 318 SOLID GEOMETRY < REVIEW. 638. State the formula for 1. The lateral area of a cylinder of revolution. 2. The total area of a cylinder of revolution. 3. The volume of a cylinder of revolution. A. The ratio of the lateral area of two similar cylin- ders. 5. The ratios of the lateral areas of any two cylinders of revolution. 6. The ratios of the volumes of two similar cylinders. 1. A protecting wall for an embankment 500' long is 12' wide at the bottom, 3' wide at the top, and 18' high. How many cubic yards of stone in the wall? 2. How many brick, 2" X 4" X 8", are required to build a chimney two bricks thick with a flue 8" X 12" and 25' high, if the mortar averages i" thick? Make a drawing of two tiers of brick when properly laid. 3. If from any point in an equilateral triangle, perpendicu- lars be drawn to the sides, the sum of these three perpendiculars equals the altitude of the triangle. 4. If from any point within a regular tetrahedron perpen- diculars be drawn to the four faces, the sum of these perpendicu- lars equals the altitude of the tetrahedron. CONES. 639. CONICAL SURFACE. A curved surface formed by a moving straight line which passes through a fixed point and continually touches a fixed curve is a conical surface. The mov- ing straight line is the generatrix and the fixed curve is the directrix. The generatrix in any one of its posi- tions is an element of the surface. Tho fixed point is the vertex of the coni- cal surface. Ft' the directrix is a closed CONES 319 curve, the conical surface is a closed surface. The por- tions of the conical surface on the two sides of the ver- tex are the nappes of the conical surface and are designated as upper and lower nappes. Usually only one nappe of a conical surface is considered. 640. A CONE. A solid bounded by one nappe of a closed conical surface and a plane cutting all the ele- ments, is a cone. The plane section is the base of the cone, The bounding portion of the conical surface is the lateral surface of the cone and the vertex of the conical surface is the vertex of the cone. The portions of the elements of the conical surface between the vertex and the base are the elements of the cone. The distance from the vertex to the plane of the base is the altitude of the cone. 641. CIRCULAR CONE. A cone with a circular base is a circular cone, and the straight line joining the vertex to the center of the base is the axis of the circular cone. 642. RIGHT CONE. A cone such that the axis is per- pendicular to the base is a right cone. 643. OBLIQUE CONE. A cone such that the axis is not perpendicular to the base is an oblique cone. 644. CONE OF REVOLUTION. A cone generated by the revolution of a right triangle about one leg as an axis is a cone of revolution. Any element of a cone of revolu- tion is its slant height. 645. SIMILAR CONES. Cones of revolution that can be generated by the revolution of similar right triangles about homologous sides are similar cones. 320 SOLID GEOMETRY 646. TANGENTS TO A CONE. A line which touches the conical surface in a point but does not intersect it is a tangent line to the cone. A plane which embraces an element but does not intersect the conical surface is a tangent plane to the cone. Lane a is tangent to the cone at the point of tangency J> and plane N is tangent to the cone along the element GH. 647. A TRUNCATED CONE. That portion of a cone in- cluded between the base and a plane cutting all the ele- ments is a truncated cone. 648. A FRUSTUM OF A CONE. A truncated cone such that the cutting plane is parallel to the base is a frus- tum of a cone. The base of the cone is the lower base of the frustum, the section is the upper base, the distance between the two parallel planes is the altitude. If the frustum be a portion of a cone of revolution, the portion of the slant height of the cone included between the base 4 of the frustum is the slant height of the frustutr PRELIMINARY THEOREMS. 649. THEOREM I. Every cone of revolution is a rigid circular cone. 650. THEOREM It. Tlic a.cis of a cone of revolution is its altitude. 651. THEOREM III. All elements of a cone of r< so- lution arc equal (i. e. the slant height r is constant to the hypotenuse of the generating triangle). CONES 321 1. Two circular cylinders have the same altitude, but the volume of one is four times that of the other. Find the ratio of their radii. PROPOSITION V. 652. THEOREM. Every section of a cone em- bracing the vertex is a triangle. Given a plane embracing the vertex A and intersect- ing the perimeter of the base in the points B and C. To Prove that the A ABC is the section made by the plane. Proof. SUG. 1. The plane must cut the perimeter in two points as B and C. 2. AB and AC are each in both the plane and the conical surface. 3. AB and AC are intersections of the plane and the conical surface. 4. What kinds of lines are AB and AC? Therefore 653. COR. I. A plane containing an element of a conical surface without being tangent intersects the surface in a second element also. 2. In each of two right circular cylinders the altitude is equal to the diameter and the volume of one is ^ that of the other. Find the ratio of the altitudes. 322 SOLID GEOMETKY i PROPOSITION VI. 654. THEOREM. Every section of a circular cone made by a plane parallel to the base is a circle. Given F OGH a circular cone with base GH and a section .O'G'H' parallel to the base, ' , G',H' being the points in which the plane intersects the ele- ments FG } FH, and the axis FO respectively. To Prove O'G'H' a O. Proof. SUG. 1. is the center of the base and G and // are taken as any two points on the perim- eter of the base. The figures FG'GOO' and FH'HOO' are plane figures. Why? 2. Compare the &FG'0' and FGO; O'C*' WTJ' FH'O' and FHO; ratios - -, - - and - FO OG Oil' Auth. 3. Compare O'G' and O'H'. 4. .'. Q' _ OH is a circle. Why? Therefore 655. COR. The axis of a circular cone passes through ike centers of all sections parallel to the base. What is the character of the polygon which revolved about one of its sides generates a frustum of a cone of revolution? COXES 323 656. CONE INSCRIBED IN A PYRA- MID. If the vertex of a pyramid coin- cides with the vertex of a cone and the base of the pyramid is circum- scribed about the base of the cone, the pyramid is circumscribed about the cone and the cone is inscribed in the pyramid 657. CONE CIRCUMSCRIBED ABOUT A PYRAMID. If the vertex of a pyra- mid coincides with the vertex of a cone and the base of the pyramid- is inscribed in the base of the cone, the pyramid is inscribed in the cone and the cone is circumscribed about the pyramid. PRELIMINARY THEOREMS AND POSTULATE. 658. THEOREM I. The faces of a pyramid circum- scribed about a cone are tangent to the cone. 659. THEOREM II. The edges of a pyramid inscribed in a cone are elements of the cone. 660. THEOREM III. The slant height of a regular pyramid circumscribed about a right circular cone equals the slant height of the cone. 661. POSTULATE I. The volume of a circumscribed pyramid, the area and perimeter of its base, and its lateral area are greater than the corresponding parts of an inscribed cone. 662. POSTULATE II. The volume of an inscribed pyramid, the area and perimeter of its base, and its lateral area are less than the corresponding parts of a circumscribed cone. 324 SOLID GEOMETRY t 663. POSTULATE III. // the number of sides of a pyramid inscribed in or circumscribed about a cone be indefinitely increased in such a manner that the faces of the pyramid be all indefinitely decreased, the volume, lateral surface, base, perimeters of the base and sections of the cone are the limits of the corresponding parts of the pyramid. State one kind of inscribed or circumscribed prisms by which the two conditions underlying the limiting process in Postu- late III may be set up in one operation. PROPOSITION VII. 664. THEOREM. The lateral area of a cone of revolution is equal to one-half the product of the perimeter of its base and its slant height. Given C BD a cone of revolution, its lateral area denoted by S, the perimeter of its base by p, and its slant height by I. To Prove S=$pxl. Proof. SUG. 1. Circumscribe about the cone a regu- lar pyramid, with lateral area denoted by S' , perimeter of the base by p'. The slant height of the pyramid is I. Why ? 2. Determine the lateral area S' in terms of p' and I. CONES 325 3. Complete the demonstration. See the method of 630. Therefore 665. COR. I. // r be the radius of the base of a cone of revolution, then the lateral area is given by the for- mula 8=\l X "2flfr=^rl. 666. COR. II. The lateral areas of two similar cones are to each other as the squares of their altitudes and as the squares of the radii (or diameters) of their bases. SUG. See method of 632. 667. COR. III. The lateral area of the frustum of a cone of revolution is equal to one-half the product of its slant height and the sum of the perimeters of its bases. SUG. If a regular pyramid be circumscribed about the cone from which the frustum is cut, the cutting plane will cut from the pyramid a frus- tum which will be circumscribed about the frus- tum of the cone. Denote the radii of the two bases by r l and r . the lateral area by S and the slant height by ?. Prove that 8 is equal to 668. COR. IV. // r represent the rad- ius of the section of a frustum of a cone of revolution midway between the bases, prove SUG. Show r= 326 SOLID GEOMETRY PROPOSITION VIII. 669. THEOREM. The volume of a cone is equal to one-third the product of its base and its alti- tude. Given the cone A FD with altitude h, base B, and volume V. To Prove V=%B x k. Sue. 1. Inscribe in the co^ie a pyramid, de- noting its base by B e , and its volume by V . Its altitude will be h. Why ? 2. Complete the proof by an adaptation of the method of Prop. VII. Therefore 670. COR. I. The volume of a circular cone is given by the formula V=^r z h. 671. COR. II. The volumes of similar cones are to each other as the cubes of the radii of their bases, as the cubes of their altitudes, and as the cubes of their slant heights. EXERCISES 327 SUG. Adapt the method of 666. 672. COR. III. The volume of a frustum of a cone of revolution is equal to one-third the product of the altitude and the sum of the upper base, the lower base, and a mean proportional between the two bases, i. e. V=h (B i +B 2 + ^'E^E z ) in which B^ andB 2 denote the two bases. SUG. Inscribe in the original cone a pyramid. The frustum of the pyramid made by the cutting plane of the cone will be inscribed in the frustum of the cone. Proceed as in Cor, III, v Prop. VII. 673. COB. IV. The volume of a frustum of a cone of revolution is given by the formula V=\*h (r l *+r z * + vTj'xT 7 ;). REVIEW. 674. State the formulas for 1. The area of a circle. 2. The lateral area of a cone of revolution. 3. The lateral area of the frustum of a cone of revolution. 4. The volume of a circular cone. 5. The ratio of the volumes of two similar cones. 6. The ratio of the lateral areas of two similar cones. 1. A granite monument is 30' high. The base is 4' in diam- eter. It diminishes gradually in size to a cross section circle 15" in diameter, 6' from the top. The remainder of the monu- ment is conical. Find its volume in cubic yards and its weight in tons, the specific gravity being 2.65. 2. How many sq. yds. of canvas in a conical tent 12$' high vith a base diameter of 12'? 3. Find the volume of the solid generated by the revolution upon one of its sides as an axis of an equilateral triangle with an edge of 6'. 328 SOLID GEOMETRY 4. Find the ratios of the volumes generated by the revolu- tion of a right triangle about side a, about side 6, and about hypotenuse o respectively. 5. The altitude of the frustum of a cone of revolution is the altitude of the cone. What is the ratio of their volumes? 6. A section of a tetrahedron cutting four edges at their mid points is a parallelogram. 7. A grain bin is 6' X 12' X 20'. How many bushels does it hold? 8. The radii of the two bases of a frustum of a cone are 4' and 9' respectively. Find the volume. Indicate and work mentally. 9. What is the locus of a rod 12' long, one end being sta- tionary on the ceiling of a room 8' high and the other end resting on the floor? 10. Find the area of the plane figure enclosed on the floor in the preceding exercise and the volume enclosed by the floor and the locus. 11. A tent has a diameter of 16' and a vertical side wall of 5'. The covering is conical in shape and the center pole is 12' high. How many cubic feet of air does the tent contain? 12. Find the lateral area, total area, altitude, and volume of the cone that can be circumscribed about a regular tetrahedron with an 8" edge. 13. A cylinder and a cone of revolution have the same altitude and concentric bases. The diameter of the base of the cone is three times that of the cylinder. How far from the vertex of the cone do the two lateral surfaces intersect? Suppose the diameter of the cylinder to be f that of the cone, where do the surfaces intersect ? 14. What is the locus of lines making a given angle with a given line at a given point in the line ? 15. The altitude of a cone is trisected by planes parallel to the base. Compare the parts into which the cone is divided. Make a similar comparison when the altitude is divided into four equal parts. 16. What kind of a triangle is the section of a right cone through the vertex? EXERCISES 320 17. What is the volume of a piece of timber 15' long, the bases being squares of 12" and 14" respectively? 18. If four similar cylinders have their altitudes proportional to 3, 4, 5, 6, prove that the volume of the largest one equals the sum of the volumes of the three others. 19. A log 20' long has a diameter at the smaller end of 16". What proportion of the log is cut into slabs if the largest possi- ble squared stick of timber is sawed from it? What proportion will be slabs if the largest rectangular stick be sawed, the edges having the ratio of 3 to 4? Of 2 to 3? 20. A cylindrical vessel with a diameter equal to its altitude holds 1414 f cu. ft. of water. What are its dimensions? 21. What are the dimensions of a cylindrical vessel holding 1414f cu. ft. if its altitude is twice its diameter? What are the dimensions if the ratio of the diameter and the altitude is f ? 22. What are the dimensions of a quart cup if the ratio of its diameter and its altitude is f? 23. What is the diameter of a cylindrical peck measure 8"" deep? 24. Allowing two inches for the seam, how large a sheet of iron will be required for a joint of 8" stove pipe? Of 6" stove pipe? 25. Measure the diameter of a quart cup and compute its depth. Verify by measurement. 26. A rectangle is 5x8. Which side as an axis will produce the greater cylinder of revolution? What is the ratio of the two cylinders? 27. Find the lateral area of a regular pentagonal pyramid with basal edge of two feet and slant height of 1 yd. 28. Find the total area of a regular quadrangular pyramid with basal edge of 12' and lateral edge of 10'. Find the vol- ume. 29. The base of a regular pyramid is a square with a side of 6' and the lateral area is f of the total area, Find the alti- tude and the slant height. 30. What is the ratio of the four parts into which a pyra- mid is divided by parallel planes dividing the altitude into four equal parts? 330 SOLID GEOMETRY 31. How much of a cube is cut off by a plane passing through the mid-points of three concurrent edges? What part is re- moved when this is done for each set of three concurrent edges! 32. If a plane is passed through the extremities of three concurrent edges of a cube, prove (1) that the tetrahedron cut off is of the cube; (2) that four such tetrahedrons can be cut off of a cube; (3) that the remainder of the cube is a regular tetrahedron equal to of the cube. Sue. Take a as the edge of the cube. Determine the bases and altitudes of the figures and compute the volumes. 33. Use the rule for finding the volume of a pyramid in order to find the volume of an 8" cube. 34. How many cubic feet are removed in boring a 6" well 180' deep? 35. How large an object at a distance of 20' will be obscured by an inch square placed 2^' from the eye? 36. How many inches from the vertex of a cone of revolu- tion 12' high must a plane be passed parallel to the base in order to bisect the cone? In order to trisect it, at what distances must the two planes be passed? 37. The interior of a rectangular bin with edges in the ratio of 2, 3, 4 has a surface of 672 sq. ft. How many bushels does it hold? THE SPHERE. 675. A SPHERE. A solid bounded by a surface all points of which are equally distant from a fixed point within is a sphere. The fixed point is the center of the sphere and the bounding surface is the surface of the sphere. 676. RADIUS OP A SPHERE. Any straight line drawn from the center of a sphere to its surface is a radius of the sphere. 677. DIAMETER OF A SPHERE. Any straight line drawn through the center of a sphere and terminated by the surface is a diameter of the sphere. 678. A LINE TANGENT TO A SPHERE. A line winch SPHERES 331 touches a sphere at one and only one point is a tangent to the sphere. 679. A PLANE TANGENT TO A SPHERE. A plane which touches a sphere at one and only one point is tangent to the sphere. 680. POINT OF TANGENCY. The point in which a tangent line or a tangent plane touches a sphere is the point of tangency or point of contact. Two spheres are tangent when they have one and only one point in common. PRELIMINARY THEOREMS. 681. THEOREM I. All radii of a sphere are equal and all diameters of a sphere are equal. 682. THEOREM II. Two spheres are equal if their radii are equal. In the case of spheres, equality implies congruence. 683. THEOREM. III. The radii of two equal spheres are equal. 684. THEOREM IV. The revolution of a semicircle about its diameter generates a sphere. 685. THEOREM V. A straight line which intersects a sphere intersects it in two points. 686. THEOREM VI. A plane or a sphere which inter- sects a sphere intersects it in a closed line. 332 SOLID GEOMETRY - / PROPOSITION IX. ; 687. THEOREM. Every section of a sphere by a plane is a circle. Given the plane N intersecting the sphere in the closed line ABD. To Prove ABD a circle. Proof. SUG. 1. Drop a 1 to plane N from the cen- ter O f meeting plane N in C. Let A and B be any two points in the perimeter of the section. Draw AC and EC. 2. Compare AC and BC. 3. -'-ABD is a circle. Why? Therefore 688. A GREAT CIRCLE OF A SPHERE. A circle of a sphere which contains the center of the sphere is a great circle of the sphere. 689. A SMALL CIRCLE OP A SPHERE. A circle of a sphere which does not contain the center is a small cir< of the sphere. 690. Axis OP A CIRCLE OP A SPHERE. The diameter of a sphere perpendicular to a circle of a sphere is the axis of the circle of the sphere. SPHERES 333 691. POLES OF A CIRCLE. The extremities of the axis of a circle are its poles. AB is a great circle, EF is a small circle, GH is the axis of Q EF and also of Q 4-B if th e two are parallel, G and H are the poles. 6C2. COR. I. The axis of a circle intersects it at its center. 693. COR. II. Two circles of a sphere equally dis- tant from the center are equal. 694. COR. III. Of two circles of a sphere unequally distant from the center, that one ivhich is nearer the center is the greater. 695. COR. IV. The center of the sphere is the cen- ter of every great circle of the sphere. 696. COR. V. All great circles of a sphere are equal. 697. COR. VI. Two great circles of the same sphere intersect in a common diameter. 698. COR. VII. Two great circles of the same sphere bisect each other, the sphere, and the surface of the sphere. 699. COR. VIII. Any three points on the' surf ace of a sphere determine a circle of the sphere. SUG. How is a plane determined? ^ 700. COR. IX. Through two points on the surface of a sphere, not the extremities of a diameter, one and 7 7 7 7 st>J /^v^"f^-*-T 0*SVf xC>-< only one great circle can be drawn. 701. DISTANCE ON A SPHERE. The distance between two points on the surface of a sphere is the shorter arc of the great circle joining them. 1. A plane embracing the axis of a circle is perpendicular to the circle. If the arc of a circle on a sphere is bisected by a plane embracing the axis, every point in the plane is equidistant from the extremities of the arc. 334 SOLID GEOMETRY i PROPOSITION X. 702. THEOREM. All points on a circle of a sphere are equally distant from each of its poles. Given ABCD, a O of sphere 0, P and P' being its poles, C and D any two points on the O, PC and PD great circle arcs. To Prove arc PC^arcPD and arc P'C = arc P'D. Proof. SUG. 1. P and C determine a great circle which also passes through P'. The same is true ofPandD. Why? 2. Draw CN and DN, radii of O ACD. 3. Compare the chords PC and PD. 4. Compare the arcs PC and PD. 5. Compare arcs PCP' and PDP' : arcs P'C and P'D. Therefore 703. POLAR DISTANCE. The distance 011 the surface of a sphere from a circle to its nearer pole is the polar distance of the circle. The polar distance of a circle is less than a great semi- circle. 704. COR. I. The polar distance of a great circle is a quadrant, i. e. an arc of ninety degrees. SUG. What angle at the center of the splinv subtends the polar distance of a great circle? SPHERES 335 PROPOSITION XI. 705. THEOREM. A point which is at the dis- tance of a quadrant from each of two points on the surface of a sphere, not the extremities of a diameter, is a pole of the great circle embracing those points. Given two points E and F on sphere 0, E and F not the extremities of a diameter, and a third point P such that arc PE= : arePF = a, quadrant. To Prove P is a pole of the great circle EF. Proof. SUG. 1. E and F determine a great circle. Why? 2. Join E and F to 0, the center of O EF, and P to 0. 3. What are the angles POE and POF1 Why? 4. What is PO with regard to O EF*>. What is P? Why? Therefore NOTE By the truth of Prop, Til arcs of great circles can be drawn on the surface of a sphere in a manner similar to that by which arcs can be drawn on a plane with given radii and centers. To do this a point is lo- cated at a quadrant's distance from the two points through which the great circle is to pass. If the divi- ders be now opened so as to span the chord subtending :]:;<; SOLID GEOMETRY ( a great circle quadrant and one point placed on the fixed point as a pole, the free end can be made to de- scribe the desired great circle. The means for deter- mining the quadrant and its chord will be found in 706. PKOPOSITION XII. 706. PROBLEM. Given a material sphere, to find its radius or its diameter. ir SUG. 1. Take any point P on the surface of the sphere as a pole and with the dividers de- scribe a circle C. 2. Take any three points A, B, D on this circle and by means of the dividers construct a AA'B'D' congruent to A ABD. 3. Determine the center C' of the cir- cle circumscribed about A A'B'D' . 4. Draw EF equal to the radius C' A' and through E draw an unlimited straight line LEF. 5. From F lay off FH = chord PB and at F erect FG _L FH, G being its intersection with EH. 6. Prove that GH is the diameter of the sphere and find the radius. 707. COR. Given a diameter of a sphere, a AB. Why ? AC = AM and hence BM>BC. Consequently M does not lie 338 SOLID GEOMETRY on the second circle and the two circles have but the one point C in common. 2. Let D and E be the points in which the two circles meet the line ADB. As A is the pole of the O CD, a line may be drawn from A to C congruent to the line AD and one can be drawn for a like reason from B to C congruent to line BE. That is, a line can be constructed from A to B through C which is shorter than the given line ADB by the line DE. Thus, no matter what line be drawn from A to B other than arc AB, C, and hence every point of arc AB } lies in a line still shorter. Therefore every point of arc AB lies in the shortest line from A to B. 3. If now D be any point not on arc AB it cannot lie on the shortest line from A to B. This is seen if the circle about A as pole and with radius AD be drawn determining a point C by its intersection with arc AB. For then, as above, there can be drawn another line from A to B through C which is shorter than any line which can be drawn through D. 4. Hence AB must be the shortest line. Therefore 709. Inasmuch as the great circle arc is the shortest line between two points on the surface of a sphere, dis- tances on a sphere are measured along great circle arcs. It will be natural then to expect on the surface of a sphere a surface or spherical geometry corresponding to plane geometry with figures formed from great circle arcs instead of straight lines. This will appear in the sequel. SPHERES 339 PROPOSITION XIV. 710. THEOREM. A plane perpendicular to a radius at its extremity is tangent to the sphere. Given sphere 0, radius OG and plane N l OG at G To Prove plane N a tangent plane to the sphere. Proof. Sue. 1. Let H be any point in plane N other than G and draw OH. 2. Where does H lie with respect to the sphere? Why? 3. Complete the demonstration. Therefore 711. COR. I. A plane tangent to a sphere is perpen- dicular to the radius at the point of contact. 712. COR. II. Any straight line through the point of tangency and in the tangent plane is tangent to the sphere. 713. COR. III. Any straight line perpendicular to a radius at its extremity is tangent to the sphere. NOTE. Of all figures in plane geometry the circle most closely resembles the sphere in its characteristics. It is interesting to note the resemblance of certain propositions and their proofs in plane and in solid geometry. The change of two words in Prop.XIV makes the theorem one of plane geometry. Compare the demon- strations of the two theorems. Such comparisons of plane and solid geometry theorems make an excellent review of much of the subject and will also in many 340 SOLID GEOMETRY instances serve to make more clear the meaning and the demon- strations of solid geometry theorems. 1. Construct a plane tangent to a sphere at a given point. 2. What is the locus of a point at a given distance from a given point and also equidistant from two other given points. Is the problem always possible? Is the locus ever a single point? PKOPOSITION XV. 714. THEOREM. The intersection of the sur- faces of two spheres is a circle. Given two spheres and 0' intersecting in a closed line, two points of which are A and B. To Prove line AB is a circle. Proof. SUG. 1. The plane AOO' intersects the two spheres in two great circles intersecting each other in the points A and C. Likewise the plane BOO' intersects the two spheres in two great cir- cles which intersect each other in the points B and D. Why ? 2. Chords AC and BD intersect 00' in the same point P. Why ? 3. Compare A AOP with A BOP. Auth. AP with BP. 4. AP and BP lie in the plane 1 00' at P. Why ? Hence the closed line AB is a circle. Therefore 715. COR. I. The line joining the centers of two in- tersecting spheres is perpendicular to the plane of their intersection and passes through the center of the circle of intersection. SPHERES 341 COB. II. The plane of the intersection of two equal spheres is the perpendicular bisector of their line of centers. 1. What is the locus of points at given distances from two given points? Discuss all possibilities. 2. Find a point X which is at given distances from two given points and equally distant from two other given points. Discuss all possibilities, showing that there are two, one, or no solutions. 3. Find a point X equidistant from two parallel lines, from two intersecting lines, and a given distance from a given point. What is the greatest number of solutions? What is the least number! Under what conditions is the problem impossible? 716. ANGLE OF Two ARCS. The angle between the tangents of two arcs at their point of intersection is the angle of the arcs. 717. SPHERICAL ANGLE. The angle formed by the arcs of two great cir- cles is a spherical angle. PA and PB are two great circles and a and 6 are tangent to the great circles re- spectively at P. The spherical angle APB equals the plane angle aPl. 718. COR. A spherical angle equals the dihedral an- gle formed by the planes of the great circles forming the spherical angle. SUG. The tangents to the great circles are in the respective planes of the circles and perpen- dicular to the edge of the dihedral at the same point. Why? 719. A CIRCUMSCRIBED SPHERE. When all the ver- tices of a polyhedron lie in the surface of a sphere, the sphere is circumscribed about the polyhedron and the polyhedron is inscribed in the sphere. 342 SOLID GEOMETRY 720. AN INSCRIBED SPHERE. When the faces of a polyhedron are all tangent to a sphere, the sphere is inscribed in the polyhedron and the polyhedron is cir- cumscribed about the sphere. PROPOSITION XVI. 721. THEOREM. One and only one sphere can be circumscribed about any tetrahedron. c Given a tetrahedron ABCD. To Prove that one and only one sphere can be circum- scribed about ABCD, i. e. to prove that there is one and only one point X equidistant from A, B } C, and D. Proof. SUG. 1. What is the locus of points equi- distant from A and #? From B and (7? Auth. 2. Show that the two loci just found must intersect. What is then the locus of points equidistant from A } B } and Cf 3. What is the locus of points equi- distant from C and D ? 4. Show that the loci in Sug. 2 and Sug. 3 intersect. 5. Show that this intersection is the required point X. 6. Show that only one such point exists. Therefore SPHERICAL POLYGONS 343 PROPOSITION XVII. 722. THEOREM. One and only one sphere can be inscribed in a tetrahedron. Given a tetrahedron ABCD. 721. To Prove that one and only one sphere can be in- scribed in ABCD. Proof. SUG. 1. What is the locus of points equi- distant f rom ABC and ABD f ABC and AC D f Auth. 2. Complete the demonstration on the outline of the demonstration 721. Therefore 1. All lines tangent to a sphere from the same point are equal and touch the sphere in a circle of the sphere. SUG. Connect the center of the sphere with the given point and with two or more points of contact. PEOPOSITION XVIII. 723. THEOREM. A spherical angle is measured by the arc of a great circle described from the vertex of the angle as a pole and intercepted be- tween the sides of the angle. Given two great circle arcs PA and PB forming a spherical angle at P, AB being an arc of that great circle of which the pole is P intercepted by AP and BP To Prove that spherical angle APB is measured by arc AB. Proof. SUG. 1. OA and OB are radii of the great circle ABC and lie respectively in the planes of 344 SOLID GEOMETRY the great circles PA and PB. What relation do they bear to POf Auth. 2. "What relation does Z AOB hear to the dihedral angle APO Bf Auth. 3. What relation does arc AB bear to AOB? ToZAP?- Auth. Therefore 724. SPHERICAL POLYGON. A portion of the surface of a sphere bounded by arcs of great circles is a spheri- cal polygon. The bounding arcs are the sides of the polygon, the intersections of the sides are the vertices, of the polygon, and the spherical angles are the angles of the polygon. The planes of the sides of a spherical polygon form a polyhedral angle with vertex at the center of the sphere. The sides of a spherical polygon are great circle arcs and may be expressed in degrees of arc. They measure the corresponding face angles of the polyhedral. The spherical angles of a spherical polygon measure the dihe- drals of the polyhedral. A diagonal of a spherical poly- gon is a great circle arc joining any two non-adjacent vertices. A diagonal of a spherical polygon is a great circle arc joining any two non-adjacent vertices. ABCD is a spherical polygon, AB is a side, A is a vertex, ABCD is the subtended polyhedral angle, AB measures L AOB, L A measures dihedral OA and AC is a diagonal. SPHERICAL POLYGONS 345 725. SPHERICAL POLYGONS classified. Spherical poly- gons, like plane polygons, are classified as triangular, quadrangular, etc., according to the number of angles or sides. They may be right-angled, isosceles, equilateral, etc., as are plane triangles. They may be congruent in that they may be applied to each other as are congruent plane triangles for two great circle arcs coincide if two points are common to them, just as is true of straight lines. They may be equal in area without being congru- ent. In congruent spherical polygons the homologous parts are respectively equal. Equal spherical angles may be made to coincide. 726. CONVEX SPHERICAL POLYGON. A spherical polygon is convex when none of its sides if extended will cut the polygon. A BCD is convex. T" 727. CONCAVE SPHERICAL POLYGONS. A spherical poly- gon which is not convex is concave or reentrant. EFGH is concave. 728. SYMMETRICAL SPHERICAL TRIANGLES. Two spher- ical triangles such that the subtended trihedral angles are symmetrical spherical trihedrals. By 508 it is seen that the homologous parts of two symmetrical spherical trihedrals are equal but arranged in reverse order. Symmetrical triangles exist in plane geometry but in that case either of them could be removed from the plane and turned over, thus reversing the order of its parts, and making the two triangles congruent. The distinc- tion of congruent and symmetric was then unnecessary. On account of the curvature of the surface of a sphere, no portion of it can be turned over and applied to any other portion. A distinction between congruent and symmetrical is in this case necessary. 346 SOLID GEOMETRY 729. OPPOSITE OR VERTICAL SPHER- ICAL POLYGONS. The polygons inter- cepted, or subtended, on the surface of a sphere by two opposite or verti- cal polyhedrals with their vertex at the center of the sphere are opposite spherical polygons. ABC and A'B'C' are opposite spherical polygons. 730. CoR.I. Opposite spherical polygons are symmetrical. 731. COR. II. Three planes not having a common line of intersection and all embracing the center of a sphere intersect the surface in two symmetrical spher- ical triangles. PROPOSITION XIX. 732. THEOREM. Two isosceles symmetrical triangles are congruent. Given two symmetrical spherical triangles ABC and A'B'C' isosceles with AB BC audA'B' =B'C'. To Prove ABC = A'B'C'. Proof. SUG. 1. Compare the face angles of the sub- tended trihedrals which are measured by AB and BC; by A 'B' and B' C' . What kind of trihe- drals are these? 2. Compare the two trihedrals. 3. Put the two trihedrals in coincidence and compare the spherical triangles. Therefore SPHERES 347 1. Demonstrate proposition 732 by superposition 733. POLAR OF A TRIANGLE. In the spherical triangle ABC, let A' be that one of the two poles of are BC which lies on the same side of BC as does the vertex A. In the same manner define B' and C" re- spective poles of arcs AB and CA. The triangle A'E'C' is the polar triangle of ABC. It is to be noted that the sides of ABC if extended will form eight spherical triangles including ABC. Of these but one answers the conditions of the definition ol* the polar. 2. Draw on a spherical blackboard or other sphere eight such spherical triangles and distinguish a triangle and its polar. PROPOSITION XX. . 734. THEOREM/ // one spherical triangle is the polar of a second, then the second triangle is the polar of the first. Given AA'B'C' the polar of A ABC. To Prove A ABC the polar of A A'B'C'. Proof. SUG. 1. "What must be proved concerning A, B, C in order to show that A ABC is the polar of A A'B'C"? 2. By 705 prove that A is a pole of B'C', that B is a pole of C'A', and that C is a pole of A'B'. 348 SOLID GEOMETRY 3. Suppose that A is not on the same side of B' C' as is A' and draw the great circle AA' . Suppose it to meet BC in X and B'C' in X' . Then both A' AX and AX' are quadrants. Why? But as AX f E + EC' = B'E + C'D - DE. 4. How many degrees in arc B'E? In arc C'D? Why? 5. Express arc DE in terms of a'. Therefore SPHERES 349 1, What is the locus of a point in space such that the sum of the squares of its distances from two fixed points equals the square of the distance between the two fixed points? PROPOSITION XXII. 737. THEOREM. Two symmetrical spherical triangles are equal. Given two symmetrical spherical triangles ABC and A'B'C' on the same, or on equal spheres. To Prove AABC = & A'B'C'. Proof. 'Sue. 1. Let P and P' be the respective poles of the small circles ABC and A'B'C', lying on the same hemisphere as the respective A. Draw the great circle arcs PA, PB, PC, P'A', P'B', P'C'. 2. Compare the chords AB, A' B' , etc. Compare the circles ABC and A'B'C' . Anth. 3. Compare the arcs PA, PB, PC, P'A', P'B', P'C'. Auth. A. Compare A PA B with AP'A'B'- &PBC with AP'B'C' &PCA with AP'C'A'. Auth. 5. If P lies within A ABC then &ABC==&PAB+&PBC+&PCA. If Plies with- out A ABC then &ABC=PAB+ APC-APCA. Similarly for A A'B'C'. 6. Compare A ABC with & A'B'C'. Therefore 350 SOLID GEOMETRY PROPOSITION XXIII. 738. THEOREM. Two triangles on the same sphere, or on equal spheres, having two sides and the included angle of one equal to two sides and the included angle of the other are either congru - ent or else symmetrical and therefore equal. c- c Given spherical triangles ABC and A'B'C' with AB=A'B', BC=B'C', and LB=LB'. To Prove A ABC = A A'B'C' when the order of ar- rangement of the respective parts is the same and A ABC symmetrical to A A 'B 'C' when the order oF arrangement is different. Proof. CASE I. SUG. Superimpose A ABC on A A'B'C'. See 725. CASE II. SUG. 1. Construct A A"B"C" sym- metrical to A ABC. 2. Compare A A"B"<" A A'B'C'. Case I. 3. Compare A ABC 728. with with A A'B'C'. Therefore 739. EQUAL POLYHEDRALS. Two polyhedrals which when placed with their vertices at the center of the same sphere intersect on the surface equal spherical poly- gons are equal polyhedrals, or ctjunl xolid angl< *. SPHERES 351 740. COR. Two trihedrals having a dihedral and the including face angles of one equal respectively to a di- hedral and the included face angles of the other are either congruent or else symmetrical and equal. SUQ. Use Prop. 738, and 718. PROPOSITION XXIV. 741. THEOREM. Two triangles on the same, or on equal, spheres having two angles and the in- cluded side of one equal to two angles and the included side respectively of the other are either congruent or else symmetrical and therefore equal. c- Given A ABC and A'B'C' on the same or on equal spheres, with L A= L A' , LELB', and arc AB = arc A'B'. To Prove A ABC and A'B'C' either congruent or else symmetrical and equal. Proof. SUG. 1. Let & EFG and E'F'G' be the re- spective polars of ABC and A'B'C' with sides e, 1, 9, e', /', 9', opposite A E, F, G, E', F', G' re- spectively. The pupil may construct the polar triangles. 2. Thene=e', /=/', LG=LG'. 3. Then by 738 A EFG and E'F'G' are either congruent or symmetrical. Homologous parts of A EFG and E'F'G' are then equal. 352 SOLID GEOMETRY 4. Since LE=LE\LV=LF and g=g', it follows that aa', &=&', and ZC=ZC". 5. Compare 4 ABC and A'B'C'. Auth. Therefore 742. COR. Two trihedrals having two dihedrals and included face angle of one equal to two dihedrals and the included face angle of the other respectively are either congruent or symmetrical. PROPOSITION XXV. 743. THEOREM. Two triangles on the same sphere, or on equal spheres, having the three sides of one equal respectively to the three sides of the other are either congruent or else symmetrical and therefore equal. Given two spherical triangles ABC- and A'B'C' on the same or equal spheres with AB = A'B', BC = B'C', CA = C'A'. To Prove A ABC and A'B'C' either congruent or else symmetrical and equal, Proof. SUG. 1. Connect the vertices with the re- spective centers and 0' . 2. Compare the face angles of the re- spective trihedrals and 0'. 3. Compare the trihedrals and 0'. 4. Compare the triangles ABC and A'B'C'. Therefore SPHERES 353 PKOPOSITION XXVI 744. THEOREM. Two triangles on the same sphere, or on equal spheres, having the three an- gles of one equal respectively to the three angles of the other are either congruent or else symmet- rical and therefore equal. Given two triangles ABC and EFG on the same or equal spheres, with LA=LE, LELF, ZC=Z with respective sides a, 1), c, c, f, g. The pupil may construct the figure from description in the text. To Prove A ABC and EFG congruent or else sym- metrical. Proof. SUG. 1. Let A A 'B' C' and E'F'G' be the re spective polars of A ABC and EFG, with respective sides a', V, c', e , /', g'. Compare a' and e' t I)' and /', c' and g'. Auth. 2. Compare A A'B'C' and E'F'G'. Auth. 3. Compare 1A', ', C" with AE',F',G' respectively. Auth. 4. Compare sides a, b, c with e, f, g respectively. Auth. 5. Compare A ABC and EFG. Auth. Therefore 745. COR. Two trihedrals having the three dihedrals of one equal to the three dihedrals of the other respec- tively are either congruent or symmetrical and there- fore equal. 1. A straight line tangent to a circle of a sphere lies in the plane which is tangent to the sphere at the point of contact. 2. What is the locus of a line tangent to a sphere at a given point? 3. By planes parallel to the base divide a pyramid into two equal parts. 354 SOLID GEOMETRY 4. If two angles of a spherical triangle are equal, the tri- angle is isosceles. Sue. Construct the polar of the given triangle. 5. The arc of a great circle drawn from the vertex of an isosceles triangle to the middle of the base is perpendicular to the base and bisects the vertex angle. 6. Determine a point X at a given distance from a fixed point, equidistant from two parallel planes, and equidistant from two given points. 7. Prove Prop. 741 by superposition. 8. If one circle of a sphere passes through the poles of a second circle of a sphere, the planes of the two circles are per- pendicular to each other. 9. Find a point X at a distance m from one point, a dis- tance n from a second point, and a distance p from a third point. V\ r hen is there no solution? When is there one? When two? Is there any other possibility? 10. The angles opposite the equal sides of an isosceles spherical triangle are equal. 11. In a given plane, find a point which is equidistant from the vertices of a triangle which is in another plane. Use locus. PKOPOSITION XXVII. 746. THEOREM. I. Each side of a spherical triangle is less than the sum of the other two. SUG. Construct the subtended trihedral angle at the center of the sphere and use 509. PKOPOSITION XXVIII. 747. THEOREM. The sum of the sides of a con- rex spherical polygon is less than 360 of arc. SUG. Construct the subtended polyhedral and use -510. 748. COR. The sum of the sides of a convex spherical polygon iff less than a great circle. SPHERES 355 749. COR. No side or diagonal of a convex spherical polygon is as great as 180 of arc. PROPOSITION XXIX. 750. THEOREM. The sum of the angles of a spherical triangle is greater than 180 and less than 540. c- Given A ABC. To Prove ZA+ZJ2 + ZO 180 and Z A + Z + Z C < 540. Proof. SUG. 1. Let kA'B'C' be the polar of &ABC. Then A = 180 a', B = l80l', C=l80c'. Why? 2. Then A + B + C = 540 - (a' + fc'+c') Why? 3. a' + &' + c' < 360. Why? 4. Hence ZA + LE + L C greater than 180andZA-fZ + ZC'< 540. Why? Therefore 1. The angles of a spherical triangle are 70, 96, and 84. How many degrees in the respective sides of the polar triangle? 2. The sides of a triangle are 90, 65, and 125. How many degrees in the respective angles of the polar triangle? 751. SPHERICAL EXCESS. A spherical triangle, unlike a plane triangle, may have two or three right angles, or two or three obtuse angles. The number of degrees in the angles of a spherical triangle in excess of two right angles is the spherical excess of the triangle. 356 SOLID GEOMETRY i 752. COR. The spherical excess of a spherical tri- angle is less than four right angles. 753. LUNE. A portion of the surface of a sphere included between two semicircles is a lune. The angle between the great circles is the angle of the lune. ABCDA is a lune, A and C are its angles and ABC, ADC are its edges. PRELIMINARY THEOREMS. 754. THEOREM I. The angle of a lune is equal to the dihedral angle of the planes of the great circles forming the lune. 755. THEOREM II. The angle of a lune is measured by the arc of a great circle described from either vertex as a pole and intercepted between the sides of the lune. 756. THEOREM III. Two lunes on the same sphere, or on equal spheres, are equal if their angles are equal. 1. What is the locus of a point which is at a given distance, a, from a given plane, and at a given distance greater than a from a given point in the given plane? 757. TRI-RECTANGULAR TRIANGLES. The eight spheri- cal triangles into which a spherical surface is divided by three great circles the planes of which are perpendicular to one another are tri-rectangidar triangles, i. e., each contains three right angles. 758. SPHERICAL DEGREE. If a tri-rectangular triangle be divided into ninety equal parts, one of these parts is a degree of surface or spherical degree. 2. One three hundred sixtieth part of the surface of a hemisphere, or one seven hundred twentieth part of the surface of a sphere/ is a spherical degree. AREA OF SPHERES 357 PEOPOSITION XXX. 759. THEOREM. The surface of a lune is to the surface of the sphere as the angle of the lune is to four right angles. Given a sphere with surface S and a lune with an- gle BAG and surface S' . S' LBAC To Prove = . 8 4 rt. 4 Proof. CASE I. L B AC commensurable with 4 rt. A. or 360. SUG. 1. From A as a pole draw the great cir- cle BC. Since the angles at A are measured by the arcs which they intercept on this great circle, arc BC and the great circle BC are commensur- able. Divide the arc BC and the circle BC by a common unit of measure and through each point of division and A pass a great circle. Compare the small lunes into which lune BAG and the spherical surface are divided. 2. Using one of these lunes as a unit compare the ratio with the ratio - 8 OBC 3. Using the angle of the unit lune as a / Ft 4 C 1 unit, compare the ratio -'- ^- with the ratio 358 SOLID GEOMETRY OBC LBAC 4. Compare the ratio : - with the ratio 360 CASE ii. L BAC not commensurable with 4 rt. or 360. SUG. Proceed as in 428. 760. COR. I. The number of spherical degrees in a lune is double the number of angular degrees in its angle. SUG. Let S denote the number of spherical de- grees in the lune and A the number of angular S A degrees in the angle. Then whence 720 360 #=2,1. 761. COR. II. The surface of a lune equals the prod- uct of a. tri-rectangular triangle by twice the number of right angles in the angle of the lune. SUG. Denote the tri-rectangular triangle by T and the number of right angles in the angle of the lune by A. Then - 4 rt. A. o 'S = 2AX T. 1. The angle of a lune is 72 and the area of the sphere is 95 sq. in. Find the area of the lune. SUG. By Cor. I, S = 2a sph. deg. = 2 x 72 sph. detfnv.~ Cor. II, S = 2 A x T = ^- T. 2. The angle of a lune is 130 and the area of the sphere is 45 sq. yds. Find the number of square yards in the lune. AREA OF SPHERES 359 PROPOSITION XXXI. 762. THEOREM. The sum of the areas of the two vertical spherical triangles formed on a hem- isphere by two intersecting great circles equals a lune with an angle equal to the angle of the inter- acting circles. Given a hemisphere ABCD with two great cir- cles ACE and BCD intersecting at C and forming two vertical spherical A ACD and BCE. To Prove AACD + A ECB = lune CAFD. Proof. SUG. Compare A ECB with its opposite, &AFD. Therefore PROPOSITION XXXII. 763. THEOREM. The number of spherical de- grees in a spherical triangle equals the number of angular degrees in its spherical excess. Given A ABC, S denoting its area in spherical de- grees, A, B, and C denoting the number of angular de- 360 SOLID GEOMETRY t grees in the angles of A ABC, and E denoting its spherical excess. To Prove 8 = E. Proof. Sue. 1. The surface of the hemisphere A BCC'B', or 360 sph. deg., equals A$ + AJ/ + AA T + &P,H,N,P and 8, denoting the number of spherical degrees in the respective triangles. 2. A 8 + A M = lune B, AS + AN = lune C, A 8 + A P = lune A. Why ? 3. Hence S + M = 2B, S + N = 2 C, S + P = 2 A, .: 2 S + 360 = 2 (A+B+C). Why ? 4. Whence S = A + B + C 180, i. e. the number of spherical degrees in A ABC equals the number of angular degrees in the tri- angle less 180, i. e. S = E. Therefore 764. COR. I. The ratio of a spherical triangle to 1? the surface of its sphere is and its ratio to a tri- pt rectangular triangle is 8 1. The area of a spherical triangle can be obtained when the area of the sphere and the angles of the triangle are known. 2. The angles of a triangle are 96, 72 and 120, and the surface of the sphere is 360 sq. in. How many sq. in. in the sur- face of the triangle? 765. SPHERICAL EXCESS OF A SPHERICAL POLYGON. The number of degrees by which the sum of the angles of a spherical polygon of n sides exceeds (n 2) 180 is the spherical excess of the polygon. AKEA OF SPHERES 361 PROPOSITION XXXIII. 766. THEOKEM. The number of spherical de- grees in a spherical polygon equals the number of angular degrees in its spherical excess. o Given a spherical polygon ABCD..., S denoting its area in spherical degrees and E its spherical excess. To Prove S = E. Proof. SUG. 1. Draw all possible diagonals from some one vertex, thus dividing the polygon in.to n 2 A. Denote their areas in sph. deg. by $ 15 S 2 , S 3 , etc., and their respective spherical excesses by E lt E 2 , E 3 , etc. = (4 of A^- 180) + (4 of AS a - 180) + . . . . = 4 of ABCD - (n - 2) 180 = E. 4. S = S l + S 2 +8 B H- . . Therefore 767. A ZONE. A portion of the surface of a sphere included between two parallel planes is a zone. The cir- cles of the sphere formed by the bounding planes are the bases of the zone and the distance between them is the altitude of the zone. 362 SOLID GEOMETRY 768. SPHERICAL SEGMENT. A portion of a sphere in- cluded between two parallel planes is a spherical seg- ment. The sections of the sphere formed by the two planes are the bases of the segment and the distance be- tween the planes is the altitude of the segment. In the figure X is a spherical segment. The spherical surface of a segment is a zone. If a portion of a sphere be cut off by a plane, this portion is a segment and its spherical surface is a zone, for they are included between the cutting plane and a tangent plane parallel to the cutting plane. In this case the segment and the zone have but one base each. Find illustra- tions of zones of both kinds from geography. 769. If a semicircle be revolved about its diameter as ;.n axis a spherical surface is generated. Any arc of the semicircle generates a zone. 770. SPHERICAL SECTOR. A portion of a sphere gen- orated by the revolution of a circular sector about a diameter is a spherical sector. It is important to form mental pictures of the different varieties of spherical sectors and de- scribe them. For example, if the semi-circle ADB is revolved about the diameter AB, the circular sector AOC generates a spherical sector the sur- face of which consists of a zone of one base gen- erated by the arc AC and a convex conical surface generated by the radius OC. The sector COD generates a spherical sector 6 bounded by a zone of two bases, a convex conical surface generated by OD, and a concave conical surface generated by OC. Generate many spherical sectors and describe them, many drawings to illustrate spherical sectors. 1. Construct a semi-circle and in it a circular sector which, if revolved, will generate a spherical sector having two concave conical surfaces. Describe the zone. AREA OF SPHERES 363 1. Construct a circular sector which generates a spherical sector the surface of which consists of a concave conical surface, a plane surface, and a zone. 2. Is a hemisphere a spherical sector? Why? A spherical segment? Why? 3. Describe the sectors that have for their spherical surface the torrid zone, the N. frigid zone, and the S. temperate zone. PROPOSITION XXXIV. 771. THEOREM. The area of the surface gen- erated by the 'revolution of a straight line seg- tnent about an axis in its plane but not crossing the axis, is equal to the product of the projection of the segment upon the axis and the circumfer- ence of a circle the radius of which is a perpen- dicular erected at the mid point of the segment and terminated by the axis. Ayjc A r 1C d N i D Given axis d with a line segment AB in the same plane as d, but not crossing it, CD the projection of AB on the axis, M the mid point of AB, MO a perpendicular to AB at M intersecting the axis at and 8 the area of the surface generated by the revolution of AB about the axis. To Prove 8 = 2* X MO x CD. Proof. CASE I. AB oblique to CD not intersect- ing it. 364 SOLID GEOMETRY SUG. 1. Draw MN 1 CD, join A and C f B and D, draw AP II CZ>. AC and Z) are 1 CD. Why ? 2. = 27r X MN X AB. Why? 3. A ABP A OMN. Why ? ' ABXMN=CDX MO, Why ? 4. ' # = 27r X MO X CD. CASE II. When AB meets CD. SUG. Follow the plan of Case I. CASE III. When AB II CD. Proof left to the pupil. Therefore 1. How many spherical degrees are there in a spherical tri- angle with angles of 200, 140, and 100? 2. What part of the surface of a sphere is a spherical tri- angle with angles of 120, 140, and 160? 772. POSTULATE. If half of a regular polygon be in- scribed in a semicircle and the number of sides be indefi- nitely increased, and if the semicircle be revolved about its diameter the surface generated by the semipolygon is a variable which approaches the sphere generated by the semicircle as its limit and the apothem of the poly- gon approaches the radius of the sphere as its limit. 3. A boiler is 4$ ft. in diameter and 18 ft. long. It is pene- trated by 48 3-inch cylindrical tubes. How many gallons does it hold! 4. A pyramid with an altitude of T 3 ff ft. is cut into two parts of equal volume by a plane parallel to the base. Find the distance of the cutting plane from the vertex. 5. A plane parallel to the base of a cone bisects the alti- tude. Compare the volumes of the two parts. 6. A circular water tank with a diameter of .10^' is 2i' deep. How many barrels will it hold? (Indicate operation and abbreviate by cancellation.) AREA OF SPHERES 365 PROPOSITION XXXV. 773. THEOREM. The area of a sphere is equal to the product of its diameter and a great circle. B Given a sphere generated by the revolution of the semicircle ACDB, its surface denoted by S, its radius by r. To Prove 8 = 2* r x 2r. Proof. SUG. 1. Divide arc ACDB into equal parts and draw the chords forming the regular semi- polygon ACDB. Draw the projections of the chords upon the diameter AB represented AC', C' V Draw the apo- them x of the polygon to each chord. 2. The surface generated by AC equals 27r X AC' X x. Why? What is the surface gen- erated by CD? By DB! 3. If 8' be the surface generated by the. semi-polygon, then 8' = 2-n- X AB X x . Why ? 4. If the number of sides be indefi- nitely increased then S' =: S, x = r 2?r AB x = 2?r AB x > ' tf =- 2,rr * 2r. Why? Therefore 774. COR. T. The .surface of a sphere equals 4irr a . or "V/-. 775. COR. II. The area of the surface of a sphere equals that of four great circles. 366 SOLID GEOMETRY t 776. COR. III. The area of the surface of a sphere equals that of a circle with a radius equal to the diam- eter of the sphere. 111. COR. IV. The surfaces of two spheres have the same ratio as the squares of their radii and the squares of their diameters. 4:7rl'~ 778. COR. V. The area of a spherical degree is 779. COR. VI. The area of a spherical triangle is . , in which b is the spherical excess of the triangles. 720 780. COR. VII. 2 lie area of a spherical polygon is -1 ., 2 V J7* , in which E is the spherical excess of the polygon. 781. COR. VIII. The area of a zone is 2-n-r x h f ; n which h is the altitude of the zone. SUG. Demonstrate according to the method of 773, using an arc less than a semicircle. 782. COR. IX. Zones on the same sphere or on equal spheres are proportional to their altitudes. 783. POSTULATE. // a polyhedron be circumscribed about a sphere and the number of its sides be indefinitely increased in such a manner that the areas of each of the faces of the polyhedron are at the same time indefinitely decreased, the surface of the polyhedron is a variable which approaches the surface of the sphere as its limit and the volume of the polyhedron is a variable which approaches the volume of the sphere as its limit. 784. SPHERICAL PYRAMID. A solid bounded by a spherical polygon and the polyhedral angle which it subtends at the center of the sphere is a spherical pyra- mid. Its vertex is the center of the sphere and its base is the spherical polygon, VOLUME OF A SPHERE 367 PKOPOSITION XXXVI. 785. THEOREM. The volume of a sphere is equal to the area of its surface multiplied by one- third its radius. Given a sphere, denoting its volume by V, its surface by S, and its radius by r. To Prove V = $rxS. Proof. SUG. 1. Circumscribe a polyhedron about the sphere and join each vertex to the center. Each face forms the base of a pyramid, the edges of which are the lines from its vertices to the center. What is the altitude of each pyramid 1 2. What is the volume of each pyra- mid ? What is the volume of the polyhedron ? 3. Apply the limiting- process of 783. Therefore 786. COB. I. The volume of a sphere i,s jTrr 3 (774), r being the radius , or JTT D*. 787. COB. II. The volumes of tu\o spheres have the same ratio as the cubes of their radii and the cubes of their diameters. 788. COB. III. The volume of a spherical sector is equal to the product of the area of its zone and one-third the radius of the sphere, i. e. $* R 2 h. SUG. Base a demonstration upon the method of 785. 789. COB. IV. The volume of a spherical pyramid equals the product of the area of its base and one third fvr 9 S the radius of the sphere, ?. c. - 720 SUG. 1. Make a demonstration based on tho method of 785. 368 SOLID GEOMETRY 2. V- 720 PROPOSITION XXXVII 790. PROBLEM. To find tlie volume of a spheri- cal segment. Given a spherical segment generated by the revolution of arc AC about the diameter OP, r and r 2 denoting the radii of its respective bases, h its altitude, r the radius of the sphere, and V the volume of the segment. CASE I. To find V in terms of r\, r 2 , and h 1} when both bases are on the same side of the center, SUG. 1. Let m represent OD, the distance from the center to the nearest base, and let Vi, V 2 , V s denote the volumes generated by the tri- angles OAB, OCD, and the sector CO A respec- tively. 2. y=F + F 1 -F a . Why? -J [27rr 2 /* + (r 2 - mTI 2 ) (m + //) - (r 2 w 2 )w], which by expanding and removing parentheses becomes = J -IT [3r 2 - 3 wi 2 - 3 A ?>i - /*. a ] // =Wi [r m 2 EXERCISES 369 3. In this last expression put the first r~ w 2 equal to r 2 2 (Auth.) and the second one equal to r- A 2 + 2hm + h 2 , for 4. .-. F=|/i[7r Therefore The volume of a spherical segment equals one-half the product of its altitude and the sum of its bases, increased by one-sixth the volume of a sphere with a diameter equal to the altitude of the segment. CASE II. When the center is between the bases. SUG. Adapt the work of case I to this case. (\\SE III. When the segment has but one base. SUG. Adapt the work of case I to this case by letting r, be zero. What term will vanish from the final formula? Translate the formula into words. Use 3} for IT in the following exercise: 1. Given a sphere 20' in diameter. Find the volume of a segment, the radii of the bases being 6 and 8 ft. respectively. 2. Given a sphere 20' in diameter. Find the volume of a segment with one base having a radius of 8'. 3. A 4-in. hole is bored through a sphere 10" in diameter. How much of the volume of the sphere is cut away? 4. Find the volume of a spherical sector having a zone with an altitude of 10" on a sphere with a radius of 20". 5. Find the volume and area of a sphere 40" in diameter. 6. A sphere is cut by parallel planes so that the diameter is divided into ten equal parts. Compare the areas of the zones and also the volumes of the spherical sectors the spherical surfaces of which are the respective zones. 370 SOLID GEOMETRY 7. If the average specific gravity of the earth is 5.6, what is its weight in tons? 8. Find the angles of an equiangular spherical triangle, the surface begin -fa that of the sphere. 9. The radius of a sphere is 3 in. and the area of a spherical triangle ABC on the sphere is 12 sq. in. The angles A and Tt are 140 and 115 respectively. Find Z C. 10. The dimensions of a rectangular parallelepiped are 3', 4'. and 12'. Find the length of a great circle of the circumscribing sphere. 11. Assume the diameters of the earth and moon to be 8,000 and 2,000 miles respectively. What is the ratio of their volumes! 12. A triangle on a 12" globe has angles of 140, 119, and 196. Compute the area. 13. The angles of a pentagonal spherical pyramid are 47, 96, 120, 142, and 87, and the diameter of the sphere is 20". Find the area of the base and the volume of the pyramid. 14. Prove that the volume of a portion of a sphere included by the planes of two great circles and the intercepted lune is to the volume of the sphere as the angle of the lune is to 4 rt. angles. Such a solid is a wedge or ungula. The lune is its base. 15. A cone of revolution and a sphere are inscribed in a cylinder of revolution. Compare the volumes of the three solids. 16. If the area of the convex surface of a right circular cone is twice the area of its base, prove that the slant height of the cone is equal to the diameter of the base. 17. If the specific gravity of an iron ball 10" in diameter is 8.1, what is its weight? 18. An iron kettle in the shape of a hemisphere has a '* Bisector ................... 11 Broken Line ............... 4 Circles ................... 73 Circle of a sphere ......... 332 Circumscribed polygons ..... 87 Composition ............... 103 Concurrent lines ........... 60 Cone ..... ................ 310 Congruent ................ 20 Cosine .................... 122 Consequents .............. 101 Constant ................. 214 Continuity ................ 134 Converse ................. 30 Cube ...................... 27(5 Curved line ............... 4 Cylinder .................. 300 Cylindrical surface ......... 300 Degree .................... 7 Depression Z of ............ 123 Diagonal .................. 53 Diagonal scale ............ 115 Diameter ................ 53 Dihedral Z ................. 248 Distance .................. 4(i Division .................. 104 Division external ....... 107. 146 Division harmonic.. ..100 Exterior Z 41 Extremes 101 Extreme and mean ratio... 145 External tangent 159 Foot of a line 227 Fourth proportional 105 Frustum, cone 320 Frustum, pyramid 290 Geometric figure 2 Geometric solid 1 Harmonic division 109 Homologous 110 Horizontal Z 123 Incommensurable* 99, 213 Indirect proof 34 Inscribed polygons 87 Inscribed prisms 295 Inscribed sphere 342 Interior angles 36 Internal division 107 Internal tangent 159 Inversion 103 Isosceles A 19 Limit of a variable 215 Line segment 3 Locus 65 Locus (in space) 234 Magnitude 2,97 Measurement 97 Means 101 Mean proportion 140 Median 47 Numerical measure 97 Oblique lines 11 Opposite interior Z 41 Obtuse A 19 Parallel lines 34 Parallel pianos 240 Parallelogram 52 376 INDEX Parallelepiped 270 Parallels postulate 34 Perigon 6 Perpendicular bisector 25 Perpendicular lines 11 Physical solid. . 1 Plane 4 Plane figure 4 Point 1 Poles 333 Polar distance 334 Polar A 347 Polygons 62 Polyhedrals 260 Polyhedrons 270 Postulate 11 Prism 271 Projection on a line 140 Projection on a plane 238 Proportion 99 Proportionals 140 Pyramids 200 Quadrilaterals 52 Quantity 97 Ratio 98 Ray 3 Regular polygons 191 Regular prisms 272 Regular pyramids 290 Right angle 7 Secant line 36,74 Sect 3 Sector . 87 Segment 87 Similar polygons 110 Similar polyhedrons 302 Similitude of polygons Ill Sine 121 Slant height (pyramid) ... .290 Slant height (cone) 320 Sphere 330 Spherical angle 341 Spherical excess 355 Spherical polygons 344 Spherical sectors 362 Spherical segments 3(52 Solid geometry 223 Straight line 2 Subtends 74 Surface I Symbols 13 Symmetrical polyhedrals. . . -260 Symmetrical spherical poly- gons 345 Tangent (circle) 83 Tangent (cone) 320 Tangent (cylinder) 310 Tangent (sphere) 330 Theorem 11 Third proportional 140 Transversal 36 Triangle 19 Trigonometric functions. .. .121 Truncated prism 275 Truncated pyramid 290 THIS BOOK IS DUE ON THE LAST DATE STAMPED BELOW AN INITIAL FINE OF 25 CENTS WILL BE ASSESSED FOR FAILURE TO RETURN THIS BOOK ON THE DATE DUE. THE PENALTY WILL INCREASE TO 5O CENTS ON THE FOURTH DAY AND TO $1.OO ON THE SEVENTH DAY OVERDUE. 4 1042? OctI3'4P l r SEP 9 OCT 14 1937 LD lii-;)< YB 1728